structure of power system

8/19/2019 STRUCTURE OF POWER SYSTEM

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EE6402-Transmission and Distri bution Department of EEE2014-2015

St. Joseph’s College of Engineering/St. Joseph’s I nstitute of Technology 1 ISO 9001:2008

UNIT I STRUCTURE OF POWER SYSTEMPART A

1. What are the components of a power system?(May 2014)

The components of power systems are Generators, Step up and Step down transformers, LoadsandTransmission lines.

2. What is meant by transmission and distribution system?

A large network which is used to deliver bulk power from power stations to the loadcenters and

large industrial consumers is called distribution system.3. What are the transmission level voltages we have in India?

Primary transmission level voltage is 132 kV, 220kV or 400kV and secondary transmission levelvoltage is 33kV or 66kV.

4. What are the various levels of generation in India?

The various levels of generating voltages are 3.3kV, 6.6kV, 11kV or 33kV.

5. What are the various levels of primary distribution in India?

The various primary distribution voltages are 11kV, 6.6kV or 3.3kV.

6. What is the usable voltage for secondary distribution?

The secondary distribution voltages are 415V & 240 V (415 volts for three phase loads and 240volts for single phase loads)

7. What is a oneline diagram?

Schematic representation of the elements of electric power system is called as one line diagram.

8. What is meant by primary and secondary transmission?

Transmission of electric power at 132kV by three phase 3 wire overhead system is known assecondary transmission.Transmission of electric power at 33kV by three phase 3 wire overheadsystem is known as secondary transmission.

9. What is meant by primary and secondary distributions?

The secondary transmission lines terminates at the substations where voltage is reduced from333kV to 11kV lines which run along the road sides of the city forms the primary distribution.A primary distribution line terminates at the distributing substations where voltage is reduced from11kV to 400 volts. Thus three phase 4 wire system which connect the distributing substation andthe consumer point forms the secondary distribution.

10.

Distinguish between a feeder and a distributor.SNO Feeder Distributor

1. Feeders are conductors or transmission lines

which carry current from the stations to thefeeding points.

Feeders terminate into distributors

2. No tapping is taken from the feeders. Distributor is also a conductor fromwhich current is tapped off for thesupply to the consumer.

3. Current carrying capacity plays a major rolein designing a feeder.

Whereas voltage drop plays a majorrole in designing a distributor.

4. Current loading remains the same along itslength.

Current loading factor varies alongits length.

11.

Define the term distributor?Distributor is a conductor from which current is tapped off for the supply to the consumers.

Feeders terminate into distributor.

12. What is a feeder? (Nov 2012)

Feeder is a conductor or transmission line which transmits current from the generating stations todifferent distributing substations.

13. Why is electrical power preferably to be transmitted at a high voltage?

Electrical power is transmitted at high voltage because 1.It reduces the volume of conductormaterial used. 2. It increases transmission efficiency.

14. What are the advantages of the HVDC transmission system over HVAC transmission?

Advantages of HVDC transmission are 1)It requires only two conductors as compared to three forac transmission.There is no inductance, capacitance, phase displacement and surge problems in

DC transmission.15. What are the demerits of HVDC transmission?


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1) Electric power cannot be generated at high DC voltages; 2)The DC voltages cannot be steppedup for transmission of power at high voltages.

16. What are the disadvantages high voltage AC transmissions?

1) An AC line requires more copper than a dc line; 2)The construction of an AC line is morecomplicated than a DC transmission line.

17. What are terminal equipment necessary in HVDC system?

The terminal equipmentnecessary in HVDC system are converters, inverters mercury arc valves,

Thyristors etc.18. Why all transmission and distribution systems are three phase systems?

A three phase A.C circuit using the same size conductors as the single phase circuit can carrythree times the power which can be carried by a single phase circuit and uses three conductors forthe two phases and one conductor for the neutral. Thus a three phase circuit is more economical

than a single phase circuit in terms of initial cost as well as the losses. Therefore all transmissionand distribution systems are three phase systems.

19. State the advantages of interconnected systems.

Any area fed from one generating station during overload hours can be fed fromanother powerstation and thus reserved capacity required is reduced, reliability ofsupply is increased andefficiency is increased.

20. Mention the limitations of using very high transmission voltage.

Limitations are the increased cost of insulation of conductors, transformers switches gears andother terminal apparatus.

21. Mention the equipment that supplies reactive power in HVDC converter stations?

AC filters, Static shunt capacitors, Synchronous condensers, StaticVAR compensators.

22. Why DC transmission is economical and preferable over AC transmission for large

distances only?

Because with larger distances, the saving in cost of DC overhead lines become greater than the

additional expenditure on terminal equipment.

23. What are the advantages of adopting EHV/UHV for transmission of ac electric power?

The advantages are reduced line losses, High transmission efficiency, Improved voltageregulation

24. Mention the problems associated with an EHV transmission?

The problems associated with EHV transmission are corona loss and radio interference,requirements of heavy supporting structures erection difficulties and insulation requirements.

25. What for series and shunt compensation provided in EHV lines?

Series compensation is provided to reduce the series reactance of the line so as to improvestability, voltage regulation and transmission efficiency. Shuntcompensation is provided to reducethe line susceptance so as to improve the voltage regulation under light load condition.

26. What is the voltage that has been selected for HVDC transmission?

The voltage level selected for HVDC transmission is±500V

27. What is FACTS?

Flexible Alternating Current Transmission System is alternating current transmission systemsincorporating power electronic based and other static controllers to enhance controllability andincrease power transfer capability.

28.

State the IEEE definition for TCSC.

A capacitive reactance compensator which consists of a series capacitor bank shunted by aThyristor-controlled reactor in order to provide smoothly variable series capacitive reactance.

29. State the IEEE definition for SVC.

A shunt connected static VAR generator or absorber whose output is adjusted to exchangecapacitive or inductive current so as to maintain or control specific parameters of the electric power system (bus voltages)

30. State the IEEE definition for STATCOM.

A static synchronous compensator is a static synchronous generator operated as a shunt connectedstatic VAR compensator whose capacitive or inductive output current can be controlled

independent of the AC system voltage.

31.

State the IEEE definition for UPFC.Unified Power Flow Controller is a combination of static synchronous compensator (STATCOM)

and a static series compensator (SSSC) which are coupled via a common DC link, to allow


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bidirectional flow of real power between the series output terminals of the SSSC and the shuntoutput terminals of the STATCOM, and are controlled to provide concurrent real and reactiveseries line compensation without an external electric energy source

32. What are the advantages of high voltage AC transmission? (Nov 2011)

1) The power can be generated at high voltages.2)The maintenance of AC substations is easy andcheaper.

34. What is ring main distributor?(Nov 2012)

In this system, primaries of distribution transformer form a loop. The loop starts from thesubstation bus bars, makes a loop through the area to be served, and returns to the substation.

35. Give the reason why transmission line are three phase three wire circuits and distribution

lines are three phase four wire circuits?(Nov 2013)

A Balanced three phase circuit does not require the neutral conductor, as the instantaneous sum of

the three line currents are zero. Therefore the transmission lines and feeders are three phase threewire circuits. The distributors are three phase 4 wire circuits because a neutral wire is necessary to

supply the single phase loads of domestic and commercial consumers.

PART B1. Give the advantages, disadvantages and applications of HVDC transmission.(Nov 2012)

Advantages, disadvantages of HVDC transmission system:

8. (ii) Advantages of EHVAC and HVDC transmission.(Nov 2013)

Applications of HVDC transmission system:The controllability of current-flow through HVDC rectifiers and inverters, their application inconnecting unsynchronized networks, and their applications in efficient submarine cables meanthat HVDC cables are often used at national boundaries for the exchange of power (in North


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America, HVDC connections divide much of Canada and the United States into several electricalregions that cross national borders, although the purpose of these connections is still to connectunsynchronized AC grids to each other). Offshore windfarms also require undersea cables, andtheir turbines are unsynchronized. In very long-distance connections between just two points, forexample power transmission from a large hydroelectric power plant at a remote site to an urbanarea, it is of great interest and several schemes of these kind were built. For interconnections toSiberia, Canada, and the Scandinavian North, it may be as result of decreased line-costs of

HVDC make also of interest, but however no such interconnection was realized as inverters areexpensive.

2. Explain how choice of voltage becomes a major factor in the line design.


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3. Give the advantages, disadvantages and applications of HVAC transmission (Nov, 2013)

Advantages of HVAC transmission system:

With increase in transmission voltage the transmission efficiency increases for a givenamount of power to be transmitted over a given distance(line losses are reduced sincethese are inversely proportional to transmission voltage).

Voltage regulation is improved because of reduction in line losses. The volume of conductor material decreases being inversely proportional to the square of

the transmission voltage. The transmission capacity of the line increases tremendously since the transmission

capacity is proportional to the square of the operating voltages.

Although the cost of tower installation and terminal equipments increase but in generaluse these costs are proportional to the voltages rather than the square of the transmissionvoltages. Consequently the overall cost of transmission voltage decreases as the voltageincreases.

With increase in voltage level installation cost of the transmission line per km decreases.

Since the surge impedance loading(SIL) is proportional to the square of the voltage(PSIL=V

2/ZO, where ZO is the surge impedance of the line) , therefore with increase of

voltage level, SIL itself increases which indicates the power transfer increase. Interconnections of the power systems in large scale is possible only with EHV

transmission. Flexibility of future system growth.

Reduction in rights of way.

Limitations and problems involved in E.H.V. Transmission:

Following are limitations or problems involved in E.H.V Transmission.

Corona loss and radio interference.

Heavy supporting structures and erection difficulties.

Insulation requirements. Suitability considerations.

Current carrying capacity.


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Ferranti effect.

Environmental and biological aspects.

Applications of E.H.V Transmission:

E.H.V Transmission is used in Electrical power distribution In cathode ray tubes to generate X-Rays and Particle beams, to demonstrate arcing,, for

ignition

In photomultiplier tubes and high power amplifier vacuum tubes and other industrial andscientific applications.

4. What are the various types of HVDC links? Explain them in detail.(Nov 2011,Nov 2012)


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5. Explain the principle of operation of compensators used for voltage control(Nov 2011)


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UNIFIED POWER FLOW CONTROLLER:

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 bidirectionalflow of real power between series output terminals of SSSC and shunt output terminals of theSTATCOM. These can be controlled to provide real and reactive series line compensation withoutan external electrical energy source. It is shown in the fig.


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6. i) Briefly discuss about static VAR compensationii) Explain break even distance.(May 2012)

(i) Static VAR compensator:

In STATCOM, converters are used while in SVC thyristors without gate turn offcapability are used. It is shunt connected static VAR generator or absorber. The output ofSVC is adjusted to control capacitive or inductive current in order to control or maintaincertain parameters normally bus voltage of the power systems. A basic model of SVC isshown in the figure. The separate equipments are present in SVC for lagging and leadingVARs. 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 squareof voltage divided by the impedance. The reactive power capability steeply falls off as afunction of square of voltage.

(ii)

Break even distance:The total capital cost of a transmission system include capital cost of substation and

capital cost of the lines including the cost of land,buildings and losses.Total capital cost of transmission system = Cost of line + capital cost of substation

=(Cost of line per km * length of line in km)+capital cost of substation

The cost per km for D.C. line is less compared to A.C line for the same power capacityand comparative reliability. The D.C. system requires only two conductors in contrast

with three phase A.C. system requiring minimum three conductors. The cost of towerinsulators and conductors of HVDC line is comparatively less than the equivalent A.C.line. A bipolar HVDC system with midpoint earthed carries same power and gives samereliability of an equivalent double circuit.The tower required for HVDC system is simpler, cheaper and easy to install than A.C.

system. The number of line conductors increases with increase in line length for A.Csystem.Below certain length of line usually 800 km, the total capital cost of HVDC is more thanA.C. line and in that case HVDC line is not preferred. This can be seen from thefollowing graph shown in the figure which gives comparison between D.C and A.C. line.


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The above graph shows the cost of power as a function of distance of line. The vertical Yaxis intersect of both the curves represent cost of terminal equipment while slope of curverepresents cost per unit length of line and cost of other accessories which changes withdistance.The curves of A.C and D.C. transmission intersect each other at appoint called break even point which indicates that after certain length of line it is preferable to use HVDC ratherthan A.C. The break even distance is different for different projects because of variations

in local conditions and cost of various equipments.The A.C. lines require intermediate substations after certain length(normally 300 km)

which is not the case with HVDC system but HVDC systems needs additional converterstations with various terminal equipments at both the ends which increases their terminalcost.

When benefit of lower line cost is more compared to higher cost of substation definitelyHVDC system becomes economical. For very long length(beyond 800 km), A.C. lines

require intermediate substations as well as intermediate compensating networks. The costof transformers used in A.C. system cannot be used beyond certain extent whereas lot ofdevelopment and progress is going on in the field of HVDC system to reduce break evendistance. The D.C lines prove to be more economical in case of long river crossings. Afterstudying technical and economical aspect for project the choice of A.C or D.C can be

made.

7. Draw the layout of modern system and explain. What is the highest voltage level available in

India for EHV transmission? (Nov 2013)

OR

Describe the basic structure of an AC power system with a single line diagram.(May 2014)

The large network of conductors between the power station and consumers can be broadly divided

into two parts viz., transmission system and distribution system. Each part can be further sub-divided into two-primary transmission and secondary transmission and primary distribution andsecondary distribution. Figure shows the layout of a typical a.c. power supply scheme by a singleline diagram. It may be noted that it is not necessary that all power schemes include all the stagesshown in the figure. For example, in a certain power scheme, there may be no secondary

transmission and in another case the scheme may be so small that there is only distribution and notransmission.(i) Generating station:

In the figure, G.S. represents the generating station where electric power is produced by3-phase alternators operating in parallel. The usual generation voltage is 11 kV. Foreconomy in the transmission of electric power, the generation voltage is stepped upto132kV at the generating station with the help of 3-phase transformers. The choice of propertransmission voltage is essentially a question of economics. Generally the primarytransmission is carried at 66 kV, 132 kV, 220 kV or 400 kV.

(ii) Primary Transmission: The electric supply (in 132kV, 220 kV, 500kV or greater) is transmit to load

center by three phase three wire overhead transmission system.(iii) Secondary transmission:

Area far from city (outskirts) which have connected with receiving station by line

is called Secondary transmission. At receiving station, the level of voltage reduced

by step-down transformers up to 132kV, 66 or 33 kV, and Electric power is

transmit by three phase three wire overhead system to different sub stations. So

this is a Secondary Transmission.

(iv) Primary Distribution: At a sub station, the level of secondary transmission voltage (132kV, 66 or 33 kV)

reduced to 11kV by step down transforms.

generally, electric supply is given to those heavy consumer which demands is 11

kV, from these lines which carries 11 kV ( in three phase three wire overheadsystem) and they make a separate sub station to control and utilize this power.


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in other cases, for heavier consumer (at large scale) their demand is about 132 kV

or 33 kV. they take electric supply from secondary transmission or primary

distribution ( in 132 kV, 66kV or 33kV) and then step down the level of voltage

by step-down transformers in their own sub station for utilization ( i.e. for electric

traction etc).

(v) Secondary Distribution: Electric power is given by (from Primary distribution line i.e.11kV) to distribution

sub station. This sub station is located near by consumers areas where the level of

voltage reduced by step down transformers 440V by Step down transformers.

These transformers called Distribution transformers, three phase four wire

system). So there is 400 Volts (Three Phase Supply System) between any two

phases and 230 Volts (Single Phase Supply) between a neutral and phase (live)

wires. Residential load (i.e. Fans, Lights, and TV etc) may be connected between

any one phase and neutral wires, while three phase load may be connected directly

to the three phase lines.

Elements of Distribution System Secondary distribution may be divided into three parts

1.

Feeders 2. Distributors

3. Service Lines or Service Mains

Feeders: Those Electric lines which connect Generating station (power station) or Sub Station to

distributors are called feeders.

Current in feeders (in each point) is constant while the level of voltage may be different; the

current flowing in the feeders depends on the size of conductor.

Distributors: Those taping which extracted for supply of electric power to the consumers or those lines,

from where consumers get electric supply is called distributors.

Current is different in each section of the distributors while voltage may be same. Theselection of distributors depends on voltage drop and may be design according voltage drop.

It is because consumers get the rated voltage according rules.

NOTE: the main difference between Feeder and Distributor is that Current in Feeder is same

(in each section) in the other hand, Voltage is same in each section of Distributor

Service Lines or Service Mains:

The Normal cable which is connected between Distributors and Consumer load terminal

called Service Line or Service Mains.

Fig. One line diagram of ac power supply scheme Fig. Elements of low voltage

distribution system.

The highest transmission voltage available as of now India is 765kV a.c. and 600kV d.c.


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8. (i) List the objectives of FACTS

(i) Objectives of FACTS:

To increase the power transfer capability of the transmission systemTo keep power flow over designated routes

9.

A 800 meters 2 wire dc distributor AB fed from both ends is uniformly loaded at the rate of 1.25A/m run. Calculate the voltage at the feeding points A and B if the minimum potential of 220 V

occurs at point C at a distance of 450 m from the end A. Resistance of each conductor is 0.05ohm/km

Solution:

Given:Current loading, i=1,25A/m

Resistance of distributor/m, r = 0.05Ω/km=2*0.05/1000=0.0001Ω Voltage at C, VC = 220VLength of the distributor, l = 800mDistance of point from A, x= 450m

Solution:

Voltage drop in section AC = irx2 /2 =

= 12.65V

Thus, voltage at feeding point A, VA = 220+12.65 = 232.65V

Voltage drop in section BC = ir(l-x)2 /2=

= 7.65V

Thus voltage at feeding point B, VB = 220+7.65 = 227.65V

10. A 2 wires dc distributor cable AB is 2 km long and supplies loads of 100 A, 150 A, 200 A and 50A situated 500 m, 1000 m, 1600 m and 2000 m from the feeding point A. Each conductor has aresistance of 0.01 ohm per 1000 m. Calculate the potential difference at each load point if a potential difference of 300 V is maintained at point A.

Solution:

Resistance of the distributor/m = 2*0.01/1000 = 0.00002Ω/m Resistance of section AC, R AC = 0.00002*500 = 0.01Ω

Resistance of section CD, R CD = 0.00002*500 = 0.01Ω Resistance of section DE, R DE = 0.00002*600 = 0.012Ω Resistance of section EB, R EB = 0.00002*400 = 0.008Ω Currents in various sections of the distributor are IEB= 50A, IDE = 50+200 = 250A, ICD = 250+150 =400A, IAC= 400+100 =500A

P.D. at load point C, VC = Voltage at A – Voltage drop in AC = VA – IACR AC = 300-(500*0.01)=295V

P.D. at load point D, VD=VC – ICDR CD = 295-(400*0.01) = 291VP.D. at load point E, VE = VD – IDER DE = 291-(250*0.012) = 288VP.D. at load point B, VB = VEB – IEBR EB = 288-(50*0.008) = 287.6V


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11. A 2 wire dc distributor 200 m long is uniformly loaded with 2 A/m. Resistance of single wire is0.3 ohm/Km. If the distributor is fed at one end, calculate (i) the voltage drop upto a distance of150 m from the feeding point. (ii) the maximum voltage drop.

Solution:

Current loading, i= 2A/mResistance of distributor per metre run, r = (2*0.3)/1000 = 0.0006Ω

Length of the distributor, l = 200m(i) Voltage drop upto a distance x metres from feeding point =

Here x = 150 m

Desired voltage drop = = 22.5V(ii) Total current entering the distributor, I = i*l = 2*200 = 400A

Total resistance of the distributor, R = r*l = 0.0006*200 = 0.12Ω

Thus, total drop over the distributor = (1/2)*I*R = (1/2)*400*0.12 = 24V

12. Explain the following (i) Stepped or tapered distributor (ii) Ring main distributor (iii) DCdistributor fed at one end. (iv) DC distributor fed at both ends. (Nov 2012)

Or13. (i) Explain the various types of DC distributors. (May 2014) OR14. Explain the following system of distribution i) radial system ii) ring main system iii)

interconnected system

1. Distributors fed at one end: In this type of feeding, the distributor is connected to supply mains at

one end and loads are tapped at different points along the path of the distributor. In this type of

distributor current in the section away from the feeding point and voltage across the loads away from

the feeding point goes on decreasing.

The minimum voltage occurs on the farthest load point. It fault occurs in any section of distributor,

the whole distributor is required to be disconnected from the supply mans and thus supply continuity

is disturbed.

2. Distributors fed at both ends: In this type of feeding, the distributor is connected to supply mains at

both ends. The voltage at both feeding points may be different or equal. In this type of distributor,

load voltage first goes on decreasing, reaches the minimum value, then starts increasing and reaches

the maximum value, when we reach the other feeding point while going from one load point to

another load point. The point of minimum voltage is never fixed. It always shifts with the variation of

load on the different sections of the distributor.

Advantages. (i) In case of fault in any one feeder feeding the distributor, the continuity of supply is

maintained by feeding it from other end.

(ii) If any section of the distributor is isolated in case of fault, the continuity of supply is maintained

to the remaining sections.

(iii) Since x-section required for doubly fed distributors is much less as compared to singly fed one,

hence it is economical.

3. Distributors Fed at the Centre: In this type of feeding, the centre of the distributor is connected to

the supply mains. In fact this type of distributor is equivalent to two singly fed distributors, each

distributor being of one half of the total length and having common feeding point.

4. Ring Mains: When the two ends of a distributor fed at equal voltages are brought together, then

such a distributor is known as ring main. It has got all the advantages of doubly fed distributor, while

the feeder required is only one. Ring main may be fed at one or more points.


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13. (ii) An electric train taking a constant current of 600A moves on a section of line betweentwo substations 8 km and maintained at 575 and 590 volts respectively. The track resistance is0.04Ω per km both go and return. Find the point of minimum potential along the track andcurrents supplied by two substations at the instant.(May 2014)

Answer : minimum potential at 2.76m from point A

Current from A is346A and from B IS 253.8A

15.

iv) design consideration in distribution system.(Nov 2013)

16. A 3 wire DC distributor is fed at one end at 220 V between wires and middle wire as shown in fig.The numbers between sections indicate the resistance of the respective section. Calculate thevoltage between middle wire and outer at each load point. (Nov 2011)

Solution:

Step1: Calculate the resistance of outer conductor and neutral conductor.Step2: Calculate the current and voltages in each section.Step3: Calculate voltage across sections AD, BE, CL, GK, FJ, HI.

UNITII TRANSMISSION LINE PARAMETERSPART A

1.

What are the primary constants of transmission lines?Resistance, inductance, capacitance and conductance distributed uniformly along the length of theline are called constants or parameters of transmission line.

2. Define resistance of transmission line?

Resistance of transmission line in a single phase is defined as the loop resistance per unit lengthof line. (Loop resistance is nothing but the sum of resistances of both the wires for unit linelength).In a three phase, it is defined as the resistance per phase. (ie) resistance of one conductor

3. Define inductance of transmission line. Give its unit.

Inductance is defined as loop inductance per unit length of line (loop inductance is the sum ofinductances of both the wires for unit line length).Its unit is henry per meter.

4.

Define capacitance of transmission line.

Capacitance is defined as shunt capacitance between the two wires per unit linelength. (or)The

capacitance between the conductors in a transmission line is the charge per unit potentialdifference.Its unit is Farad per meter.

5. What is skin effect? Is it applicable to DC current also?(Nov 2012, May 2014).


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Let A,B,Cetc be the group of conductors operating at potentials such that charges QA ,QB,Qcetc.

coulomb per meter length. ......

2d

1lncQ

1d

1ln

BQ

r

1ln

AQ

o2Π

1

AV

Where r-radius of

the conductor A. d1,d2…-distance between the conductor A and other conductor B,C etc., o-Permittivity of free space.

16. Define proximity effect on conductors.(May 2014)

The alternating magnetic flux in a conductor caused by the current flowing in a neighboringconductor gives rise to circulating currents which cause an apparent increase in the resistance of aconductor. This phenomenon is called proximity effect.

17. What is the effect of proximity effect?

Proximity effect results in i) thenon-uniform distribution of current in the cross section.ii)Theincrease of resistance.

18. What is ACSR conductor?

ACSR conductor is an aluminum conductor with a steel core reinforced. It consists of central coreof galvanized steel strand surrounded by a number of aluminum strands.ACSR is a compositeconductor which combines the lightness, electrical conductivity and rustleness of aluminum withthe high tensile strength and has a larger diameter. So to minimize the corona losses they are nowused as overhead conductors in the long distance transmission lines.

19.

What is a composite conductor?

A conductor which operates at high voltages and composes of two or more elements or strands,electrically in parallel is called as a composite conductor.

20. What is bundle conductor?

A bundle conductor is a conductor made up of two or more sub conductors and is used as one phase conductors.

21. What are the advantages of using bundled conductors?

The advantages of using bundled conductors are reduced reactance, reduced voltage gradient.

22. Define symmetrical spacing.

In three phase system when the line conductors are equidistant fromeach other then it is calledsymmetrical spacing.

23.

What is the need of transposition? (Nov 2011)When three phase line conductors have unsymmetrical spacing the flux linkages and inductances

of each phase are not the same. This results in the unequal voltage drops in the three phases evenif the currents in the conductors are balanced. Therefore the voltage at the receiving end will not

be the same for all phases. To avoid the unbalancing effect the positions of the line conductors areinterchanged at regular intervals along the line so that each conductor occupies the original position of every other conductor over an equal distance. This exchanging of positions ofconductors is called transposition.

24. Write the expression for a capacitance of a single-phase transmission line. (Nov 2012)

Capacitance per unit length between the conductors

.F/mln

0πε

ABC r D

Capacitance between line and neutral conductors

F/m.r Dln

02πAN

C

25. Define the term critical disruptive voltage?(Nov 2011)(Nov 2013)

The potential difference between conductors, at which the electric field intensity at the surface ofthe conductor exceeds the critical value and corona occurs is known as critical disruptive voltage.

26. A three phase transmission line has its conductor at the corners of an equilateral triangle

with side 3m. The diameter of each conductor is 1.63cm. Find the inductance per km per

phase of the line. (Nov 2013)

L = 0.22mH/kmr

d2lnμ10 r

7

PART B1. Derive an expression for the capacitance of a three phase line with equilateral spacing.


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2. Derive the capacitance of single phase two wire line taking earth’s effect into account.


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3. Derive the expression for the inductance of each line when the conductors are

unsymmetricalplaced.


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4. Explain about skin effect and proximity effect.(Nov 2013)

The distribution of current throughout the cross-section of the conductor is uniform when onlyD.C. is passing through it. On the contrary when AC is flowing through the conductor, the current

is non-uniformly distributed over the cross-section in the manner that the current density is higherat the surface of the conductor compared to the current density at its centre. This effect is more pronounced as frequency is increased. This phenomenon is called skin effect. It produces large


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6. Explain in detail the theory of corona formation. Give its advantages and disadvantages. And theexpression of power loss. (Nov 2012) or

7. Explain the flowing with respect to corona (i) corona (ii) effects (iii) disruptive critical voltage(iv) visual critical voltage (v) corona power loss.(Nov 2012)


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8. (i) Explain the various factors affecting the corona loss. (Nov 2011) (ii) Derive the expression of induced voltage in communication line due to current in power

lines. OR(Explain about interference between power and communication circuits (Nov 2013)


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(i)

Factors affecting corona:(a) Atmosphere: As corona is formed due to ionization of air surrounding the conductors,

therefore it is affected by the physical state of atmosphere. In the stormy weather the


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12. Find the inductance per phase per km of double circuit three phase line shown in fig. The line iscompletely transposed and operated at a frequency of 50 Hz. r =6mm.(Nov 2011)

Solution:

Inductance of each conductor = √ Where

=2.73μH 13. Find the capacitance between the conductors of a single phase 10 Km long line. The diameter of

each conductor is 1.213cm. The spacing between the conductors is 1.25m.(May 2012) Solution:

=5.99pF14. Derive an expression for capacitance of three phase unsymmetrically spaced transmission

line.(Nov 2012,Nov 2013)

The calculation of capacitance in case of conductors in three phase system which are not equally

spaced is difficult. If the line is untransposed the capacitances of each phase to neutral is not same. In

case of transposed line the average capacitance of each line to neutral over a complete transposition

cycle is same as the average capacitance to neutral of any other phase. Each conductor occupies the

same position of every other conductor after equal distance. The effect of unsymmetry between the

lines is small and calculations are carried out by considering transposition of lines.

The Fig. 1 shows three phase line with unsymmetrical spacing. The radius of each conductor is r.

When phase 'a' is in position 1, 'b' in position 2 and 'c' is in position 3

When phase 'a' is in position 2, 'b' in position 3 and 'c' is in position 1.

When phase 'a' is in position 3, 'b' in position 1 and 'c' is in position 2.


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Average voltage between conductors 'a' and 'b' is given by

Similarly average voltage drop between a and c is given by,

We have, Vab + Vac = 3 Van

For balanced circuit (qa + qb + qc ) = 0

UNIT III MODELLING AND PERFORMANCE OF TRANSMISSION LINESPART A

1. Give the lengthwise classification of transmission lines.

Transmission lines are classified as short transmission lines (length


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percentage of receiving end voltage. Χ100.

R V

R V'R V

n%Regulatio

Where'

RV - no load voltage at

the receiving end, VR - receiving end voltage.

3. Define efficiency of a transmission line.

Efficiency of a transmission line is defined as the ratio of power received to the power sent.

Χ100.sCosφsIsV

R CosφR IR V

Χ100outsentPower

deliveredPower

η Where VR ,IR ,CosR are the receiving end voltage,

current and power factor respectively.Vs, Is, Coss are the sending end voltage, current and powerfactor respectively.

4. Explain the influence of power factor on the regulation of a transmission line.

1) When the load power factor (cosR ) is lagging or unity or leading that IRcosR > IXL sin R then voltage regulation is positive (receiving end voltage is lesser than the sending end voltage)and increases with the decrease in power factor for lagging loads (for a given VR and I.

2) When the load PF is leading to this extent that IR cosR < IXL sin R the voltage regulation isnegative and decreases with the decrease in power factor for leading loads (for a given VR and I)

5. Under what circumstances, the receiving end voltage may be higher than that of the sending

end?

When load power factorcosR is leading, IXL sin R >IR cosR then regulation is negative (i.e.).The receiving end voltage may be higher than that of the sending end.Where I – load current,XL -

loop reactance,cosR - receiving end power factor(leading)

6. Explain how capacitance effects are taken into account in medium transmission lines.

Medium transmission lines have sufficient length (80-250km) and operate at voltages greater than20kV. In such lines the capacitive current is appreciable and hence cannot be neglected. So to

obtain reasonable accuracy the effects of capacitance must be taken into account.

7. What are the methods that are used for obtaining the performance calculations of medium

lines?

The methods that are used for obtaining the performance calculation of medium lines areend

condenser method, nominal T method, nominal method

8.

What is the difference between nominal T and nominal configuration?(MAY 2014) S.NO Nominal T Nominal

1) In this the whole line capacitance isassumed to be concentrated at the middle point of the line and half the lineresistance and reactance are lumped onits either side

In this the whole line capacitance isassumed to be divided into two halves,one half being connected at thereceiving end and other half at thereceiving end.

2)Full charging current flows over half theline

Capacitance at the receiving end has noeffect on the line drop. But the chargingcurrent of the second half capacitance is

added to obtain the total sending current

3) T-equivalent circuit -equivalent circuit

9. What are the limitations of nominal T and methods in transmission lines problems?

Generally the capacitance is uniformly distributed over the entire length of the line. But for easy

calculations in nominal T and the capacitance is concentrated at one or two points also in

nominal method the capacitance connected in the load side has no effect on voltage drop. Dueto all these there may be considerable error in calculation.

10. How the capacitance effects are taken into account in a long transmissionline?

Long transmission lines have sufficient length and operate at voltage higher than 100kV theeffects of capacitance cannot be neglected. Therefore in order to obtain reasonable accuracy in

long transmission lines calculations, the capacitance effects must be taken into account.

11. What is surge impedance?

The square root of the ration of line impedance(Z) and shunt admittance(Y) is called the surge

impedance(Z) of the line. 12. Define surge impedance loading or natural power of the line?


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Surge impedance loading is defined as the load of unity power factor that can be delivered by the

line of negligible resistance.oZ

2RLV

SILP Where2

RLV -line voltage at the receiving end, Zo-surge

impedance in ohms, PSIL-surge impedance loading.

13. What are the ABCD constants?

ABCD constants are generalized circuit constants of a transmission line. They are usually

complex numbers. Input voltage and current are expressed in terms of output voltage andcurrent. The constants A and D are dimensionless B and C is ohms and mhos respectively.

14. What is a power circle diagram?

A power circle diagram is a diagram drawn for the transmission line network involving thegeneralized circuit constants and the sending end voltage VS and receiving end voltage VR .

15. What is the use of power circle diagram?

Power circle diagram is used to determine the maximum power that can be transmitted over theline both at the receiving end and sending end.

16. Define attenuation in a transmission lines?

Attenuation is defined as the power loss in line. It is nothing but the transmission loss (i.e.). Thedifference between the sending end power and receiving end power.

17. Define visual critical voltages

Visual critical voltage is defined as the min. phase neutral voltage at which corona glow appearsall along the line conductors

18. Write an expression for the power loss due to corona.

5X102cVVd

r

δ

25f 242.2P

kW/km/ph. Where f - supply frequency Hz, V – phase to

neutral r.m.s voltage in kV, Vc – critical disruptivevoltage (r.m.s) per phase.

19. What are the units for A,B,C and D in the ABCD parameters?

A and D are dimensionless B and C are ohms and mhos respectively.

20. What are the voltages regulating equipments used in transmission system?

Synchronous motors, tap changing transformers, series shut capacitors, booster transformers,compound generators, induction regulator.

21.

What are the methods used for voltage control of lines?The methods used for voltage control of lines arei) by using over compound generator ii) byexcitation control

22. Distinguish between attenuation and phase constant.(Nov 2011)

,cZimpedancesticsCharacteri Y Z jβαZYconstantγnPropagatio .α – Attenuation

constant, β – phase constant

23.What is Ferranti effect?(Nov 2011)(Nov 2013)

The phenomenon of rise in voltage at the receiving end of the lightly loaded or unloaded line

is called as Ferranti’s effect.

24. What are the voltages regulating equipment used in transmission system?

a) Synchronous motors; b) Tap changing transformers.

25. Mention the methods used for voltage control of lines.a) Tap changing auto- transformer; b) Booster transformer.

26. What is sending end power circle diagram?

The circle drawn with sending end true and reactive power as the horizontal and verticalco-

ordinates are called sending end power circle diagram.

27. What is receiving end power circle diagram?

The circle drawn with receiving end values are called receiving end power circle diagram.

28. What is the range of surge impedance ina underground cable?(Nov 2012)

The range of surge impedance in a underground cable is 40 to 60Ω

PART B

1. Deduce the expression for (a) %regulation (b) ABCD parameters of a short transmission line.


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3.

Deduce the expression for (a) %regulation (b) ABCD parameters of a medium transmissionline represented in nominal T configuration.


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4. A balanced three phase load of 30MW is supplied at 132kV, 50Hz and 0.85p.f. lagging bymeans of a transmission line. The series impedance of a single conductor is (20+j52)Ω andthe total phase-neutral admittance is 315x10-6 Siemen. Using nominal T method, Determinei) A, B, C and D constants of the line (ii) sending end voltage (iii) regulation of the line. (Nov

2011)


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5. A 132kV, 50Hz, three phase transmission line delivers a load of 50MW at 0.8p.f. lagging at

the receiving end. The generalized constants of the transmission line are: A = D = 0.951.4o ;

B = 9678o; C = 0.001590

o. Find the regulation of the line and charging current. Use

nominal T method.


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6. A three phase, 50 Hz, 150 km line has a resistance, inductive reactance and capacitive shuntadmittance of 0.1 ohm and 3 x 10 -6S per km per phase. If the line delivers 50 MW at 110 kVand 0.8 power factor lagging, determine the sending end voltage and current. Assume anominal π circuit for the line.


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7. A three phase line having an impedance of (5 + j20) ohms per phase deliver a load of 30MW at a power factor of 0.8 lag and voltage of 33 kV. Determine the capacity of the phase

modifier to be installed at the receiving end if the voltage at the sending end is to bemaintained at 33 kV. Assume the shunt admittance is neglected. (Nov 2011)


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0

0

0

( )

2

5 20 20.61 75.96

Y=0 (neglected)

1A= 1+ 1 0

2

20.6175.96

75.96

33/ 3 19.05

17.61

17.61

' 15.2 ( )

capacity of the phase modifier for 3phas

r ph s

s r

r

Z R jX j

Assume

YZ D

B Z

V kV V

V V MW

B

AV MW

B

MM MVA perphase

e=15.2 3 =45.6 MVA

8.

A 15Km long three phase overhead line delivers 5MW at 11kV at 0.8 lagging power factorline loss is 12% of power delivered. Line inductance is 1.1mH per km per phase. Find

sending end voltage and voltage regulation. (Nov 2012)

9. Explain the real and reactive power flow in transmission line.

Let,


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\

10. (i) Perform the analysis of long transmission lines using RIGOROUS method.

In case of long transmission lines, for accurate solutions, the parameters must be taken as

distributed uniformly along the length as a result of which the voltages and currents will vary

point to point in the line. Given below is the equivalent circuit of the long transmission linefor rigorous method of analysis of long transmission lines. Figure shows one phase of a

distributed line of length l km.


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The real part α is known as the attenuation constant, and the imaginary part β is known as the

phase constant. β is measured in radian per unit length.


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Substituting the values of and in the equations of and we get,

The equations of voltage and current are rearranged as follows:


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Our interest is in the relation between the sending end and receiving end of the line.

Therefore, when x=l, V(l)=

, and I(l) =

. The result is,

Making use of the idendity in the above equatios we have,

(ii) Explain the concept of surge impedance loading. (Nov 2012) (Nov 2013)


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When the transmission line is loaded by being terminated with impedance equal to itscharacteristic impedance, the receiving end current is,

For a loss less line, is purely resistive. Surge impedance loading (SIL) of a transmission line isdefined as the power delivered by a line to purely resistive load equal in value to the surge

impedance of the line. SIL is given by,

11. Explain the method of drawing receiving end power circle diagrams. (May 2014)


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12. A 50Hz, three phase transmission line is 250 Km long. It has a total series impedance of (40+j100)

ohms and a shunt admittance of 914 x ohms. It delivers 50MW at 220 kV with a powerfactor of 0.9 lag. Find the Sending end voltage, Voltage Regulation, Transmission efficiency bynominal - T method. (May 2014)

Similar to problem 5

13. A three phase, 50 Hz transmission line, 40 Km long delivers 36 MW at 0.8 power factor lagging at60kV (phase). The line constants per conductor are, R = 2.5Ω, L = 0.1H, C = 0.25µF. Shunt leakage may be neglected. Determine the voltage, current, power factor, active power and reactivevoltamperes at the sending end. Also determine the efficiency and regulation of the line using

nominal π method. (Nov, 2013)

Similar to problem 6

UNITIV INSULATORS AND CABLESPART A

1. Where polythene cables are used?

i) Non-hygroscopic used in cables for submarines and damp soil.ii) Lighter used as aerial cables

for vertical installations.

2. State the advantages of polythene insulators. Non-hygroscopic, light in weight, Low dielectric constant, Low loss factor and Low thermalresistance.

3. By what materials cable sheaths are made?

Lead sheaths and Aluminum sheaths.

4. In what way Aluminium sheaths are superior to lead sheaths?

Aluminum sheaths are smaller in weight, high mechanical strength, greater conductivity, cheap,easy to manufacture and install, withstand the required gas pressure without reinforcement.

5. Where corrugated seamless aluminium sheath is used in cables?

Corrugated seamless aluminium (CSA) sheath is used in high voltage oil filled cables andtelephone lines.

6. Why corrugated seamless aluminium sheath is used in cables?

It is used because it is very flexible and easily by repeated bending the sheath isnot distorted and itis not damaged. It has lesser weight and reduced thickness.

7. Why protective covering is done in cables?


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To protect the cables from mechanical damage, corrosion and electrolytic actionwhen laid direct inthe ground the protective covering is made.

8. By what material protective covering is made in cables?

1. Bitumen&Bituminized materials, 2. PVC and 3. Layers of fibrous materials.

9. What is meant by serving of a cable?

Layers of fibrous material permitted with waterproof compound applied tothe exterior of the cableis called serving of a cable.

10. Whyarmouring is done in the cables?To protect the sheath from mechanical damage.

11. Whyarmouring is not done in single core cables?

The presence of magnetic material within the alternating magnetic field of asingle core cable produces excessive losses. Hence single core cables are leftunarmoured with non-magnetic

materials like tin-bronze or silicon-bronze tapesor wires.

12. WhyAluminumis used as an armour material?

It has non-magnetic properties, high conductivity and mechanical strength.13. What is meant by grading of cables? (Nov 2012)

The method of equalizing the stress in the dielectric of the cable is called thegrading of cables.

14. Why the capacitance of the cable is very high than the capacitance of the overheadlines?

The distances between the conductors are small. The distance between thecores and the earthed

sheath is also small. The permittivity of the cableinsulation is 3 to 5 times greater than that of airinsulation.

15. Write the expression of the capacitance of a single core cable.

C = (2πε0εr ) / ln ( R/r ) F/m, Where,R = Resistance of a conductor, r = Radius of a conductor, ε0 = 8.854 x 10

-12 F/m, εr = Relative permittivity of the cable insulation.

16. What is meant by charging current of a cable?

The capacitance of a cable determines the charging current. The charging currentrestricts the use

of cables on EHV lines. The current carrying capacity of an AC cable is also reduced by thecharging current.

17. Why power loss occurs in the dielectric of a cable?

1. Due to conductivity of insulation2. Dielectricabsorption.3. Ionization or corona.

18. Mention the methods of laying the cables.

Direct laying, draw in system and solid system.19. Mention the advantages of direct laying of cables.

It is simple and cheaper method. It gives the best conditions fordissipating the heat generated inthe cables.

20. State any two disadvantages of direct laying method.

1. Localization of fault is difficult. 2. It cannot be used in congested areaswhere excavation isinconvenient.

21. Mention the disadvantages of pressure cables.

The cost of the pressure tube is high.

22. Mention the types of gas pressure cables.

External and internal pressure cables.

23. What are the types of oil filled cables?

1. Singlecore conductor channel cables2. Sheath channel cables and3. Threecore filler spacechannel.

24. What are the types of pressure cables?

1. Oil filled cables.2. Gas pressure cables.

25. What is the operating range of pressure cables?

It is greater than 66 kV.

26. What are the advantages of separate lead screened (SL) cables over H- type cables?

The possibility of core to core breakdown decreases to a large extent.Bending of cables becomes

easy owing to no overall lead sheath.27. Mention the disadvantages of oil filled cables.

1. Expensive, 2. Laying and maintenance of cables is quite complicated.

28. What are the types of screened cables?H type and Separate Lead screened (SL) type cables.

29. Why the working voltage level of belted cables is limited to 22 kV?


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It is limited because beyond 22 kV tangential stresses acting alongthe layers of paper insulationset up large current. These current causes local heating resulting in the risk of breakdowninsulation atany moment.

30. Up to what voltage range are belted cables used?

Upto 11 kV. In some extra ordinary cases they are used upto even 22kV.

31. What are the different types of cables that are generally used for 3 phase service?

1. Belted cable, 2. Screened cables and 3.Pressure cables.

32. Why cables are not used for long distance transmission?Cables are not used for long distance transmissions due to their large chargingcurrents.

33. Mention the 3 main parts of the cable?

The main parts of the cable are Conductor,Dielectric,and Sheath.

34. What is the function of conductor?

Conductor provides the conducting path for the current.

35. What is the purpose of insulation in a cable?

The insulation or dielectric withstands the service voltage and isolates theconductor with otherobjects.

36. What is the function of sheath in a cable?

The sheath does not allow the moisture to enter and protects the cablefrom all external influenceslike chemical or electrochemical attack fireetc.

37. Mention the conductor materials in cables?The materials used for the conductor are Copper, Aluminum.

38. What is the purpose of stranding of conductors?

Stranding increases the resistance of the cable. It has flexibility.

39. Define the segmental conductors.

The stranded wires which are compacted by the rollers to minimize the air spacesbetween theindividual wires are called segmented conductors .Here the conductorsize is reduced for a given

conductance.

40.State the properties of insulating materials.

It should have high insulation resistance,high dielectric strength,and goodmechanical properties,non-hygroscopic, capable of being operated at hightemperatures, low thermalresistance and low power factor.

41. Mention the commonly used power cables.Commonly used power cables are Impregnated paper, Polyvinyl chloride, Polyethene.

42. Mention the advantages of PVC over paper insulated cables.

The advantages of PVC are reduced cost and weight, Insulation is resistant to water, simplified jointing, increased flexibility.

43. State the merits of paper insulated cables.

The merits of paper insulated cables are high current carrying capacity, long life andgreaterreliability.

44. What are the advantages of string insulators?(Nov 2011)

i) Number of units can be increased. ii) Replacement of fault insulator unit is possible.iii) Lowtension due to its swinging

45. What are the methods of grading of cables? (Nov 2011)

The methods of grading of cables are Capacitance grading and Inter sheath grading.

46. Give the relation for the insulation resistance of a cable (Nov 2013)

R=ρl/a, ρ= resistivity, l= length of the cable, a= area.

47. What is shackle insulator?(May 2014)

Shackle Insulators are frequently used for low voltage distribution line. It can be used either in ahorizontal or in a vertical position. They can be directly fixed to the pole with a bolt or to thecross arm. The conductor in the groove is fixed with a soft binding wire.

48.What is meant by dielectric stress in a cable? (May 2014)

Under operating conditions, the insulation of a cable is subjected to electrostatic forces. This isknown as dielectric stress. The dielectric stress at any point in a cable is in fact the potential

gradient (or Electric field intensity) at that point.

49.What are the different types of overvoltage tests on insulators?i) Power frequency voltage tests, ii)impulse voltage tests.

50. Define creepage distance


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It is the shortest distance on the contour of the external surface of the insulator unit orbetween twometal fittings on the insulator.

51.Define disruptive discharge voltage.

This is defined as the voltage which produces the loss of dielectric strength of insulation. It is thatvoltage at which the electrical stress in the insulation causes afailure which includes the collapseof voltage and passage of current.

52. How does the grading improve the string efficiency?(Nov, 2013)

In this method, Insulators of different dimensions are so chosen that each has a differentcapacitance. The insulators are capacitance graded i.e they are assembled in this string in such a

way that the top unit has minimum capacitance, increasing progressively as the bottom unit isreached. Since voltage is inversely proportional to capacitance, this method tends to equalize the potential distribution across the units in the string.

PART B

1. Describe the general construction of an underground cable with the neat diagram. (Nov 2012)

Explanation for various parts such as cores or conductor, lead sheath, armoring, bedding,serving, paper insulation

2. Derive the expression for the insulation resistance of a single core cable.Diagram and derivation for the insulation resistance

3. Deduce an expression for capacitance of a single core cable.

4. Discuss grading of underground cables(May 2014)

The process of achieving uniform electrostatic stress in dielectric of the cable is calledgrading of cable. Various types of grading are capacitance grading and intersheath grading.Capacitance grading: The process of achieving uniform dielectric stress by using differentdielectric is known as capacitance grading.Intersheath grading: In this method of grading, a homogeneious dielectric is used, but it is

divided into various layers by placing metallic intersheath between the core and lead sheath.Expalnation for each method.

5. Draw and explain the pin and suspension type insulators.Refer question 12

6. A single core cable has a conductor diameter of 1 cm and insulation thickness of 0.4 cm. If the

specific resistance of insulation is 5X1014 ohm-cm.Calculate the insulation resistance for a 2 Kmlength of the cable.


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7. In a three phase overhead system each line is suspended by a string of 3 insulators. The voltagesacross top unit and middle unit are 8kV& 11kV respectively. Calculate a) the ratio of capacitance

between pin and earth to the self capacitance of the each unit. b) the line voltage c) stringefficiency.(Nov 2011)

8. With neat diagrams explain constructional features of various types of cables.(Nov 2011, Nov

2012)

Various types of cables are belted type cable (upto 11KV),

Screened cable(22KV to 66KV), and pressure cable(beyond 66KV).Diagram and explanation of each cable

9. The insulation resistance of the single core cable is 495 Mega-ohms per km. If the core diameteris 2.5 cm and the resistivity of insulation is 4.5 x 10

14 ohm-cm. Find the insulation thickness.

10. Explain in detail the different methods of improving the string efficiency. (Nov 2012) Methods of improving the string efficiency: Longer cross arms, by grading the insulators, byusing guard ring.

Explanation of each type11. i) Why are insulators used with overhead lines? Discuss the desirable properties of insulators.

ii) An insulator string for 66kV lines has 4 discs. The shunt capacitance between each joint andmetal work is 10% of the capacitance of each disc. Find the voltage across the different disc andstring efficiency.(Nov 2013)


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12. (i) Describe on experiment to determine capacitance of belted cable.


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(iii) A 33kV single core cable has a conductor diameter of 1cm and a sheath of insidediameter 4 cm. Find the min and max stress in the insulator .(Nov 2013)

13. Draw with a neat sketches and explanation of pin and suspension type insulators.Compare theirmerits and demerits.

14. Elaborate the different types of Power Frequency Tests on insulators.

Power Frequency Test


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15. Elaborate the different types of impulse tests on insulators.1.Power Frequency Test : Dry and wet flashover test, Wet and dry withstand test,2.Impulse test :Impulse withstand voltage test, Impulse flashover test,3.Pollution test

UNITVMECHANICAL DESIGN OF LINES AND GROUNDING

PART A1. What is a substation?

The assembly of apparatus used to change some characteristic (e.g.: voltage, AC to DC,frequency, power factor etc) of electric supply is called a substation.

2. Mention any two layouts of laying out a substation.

a) Location should be at the center of the load, b) should provide safe and reliable arrangement.3. How substations are classified?

a) Service requirement, b) Constructional feature.

4. State the various types of substation according to its service requirements.

a) Transformer substation, b) Switching substation.5. List the types of substations classified according to its construction.

a) Indoor substation, b) Outdoor substation, c) Pole mounted substation.

6. Mention any two comparisons between indoor and outdoor substations.

Indoor : Space required and clearances between the conductors are less.Time required for erection

and possibility of faults are more.Outdoor : Space required and clearances between conductors are more.Time required for erection

and possibility of faults are less.

7. What are the major equipments of a substation?(May 2014)a) Transformer, b) Busbars, c) Insulators.

8. Define step potential.

It is the voltage between the feet of a person standing on the floor of the substation with 0.5m

spacing between two feet during the flow of earth fault current through the earthing system.

9. Define touch potential.

It is the voltage between the fingers of raised hand touching the faulted structure and the feet ofthe person standing on the substation floor. The person should not get shocked even if the earthstructure is carrying faulted current .i.e touch potential should be very low.

10. Mention the factors that affect sag in the transmission line.

a) Weight of the conductor, b) Length of the span

12. What is the reason for the sag in the transmission line?

While erecting the line, if the conductors are stretched too much between supports then there prevails an excessive tension on the line which may break the conductor. In order to have safetension in the conductor a sag in the line is allowed.


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13. What is neutral grounding?

Neutral grounding is connecting the neutral or star point of any electrical equipment (generator,transformeretc) to earth.

14. Define coefficient of earthing.

It is defined as the ratio of highest rms voltage of healthy line to earth to line to line rms voltage.

15. Define resonant frequency

It is defined as a reactance earthing with selected value of reactance to match with the line to

ground capacitance.16. Mention the disadvantages of ungrounded neutral

Occurrence of insulation breakdown leading to the heavy phase to phase fault condition.Voltagesdue to lightning surges do not find path to earth.

17. Mention two advantages of neutral grounding.

a) Arcing grounds are eliminated.b) The over voltages due to lightning and switching surges aredischarged to ground.

18. Enumerate the various methods of neutral grounding.(May 2014).

a)Solid groundingb) Resistance grounding.

19. Give the response of resistance for earth driven rods.

R= ρ /2πl *ln (2l/d), Where,l = length of the rod, d = diameter of the rod, ρ =resistivity of the rod.

20. What are the various methods of earthing in substation? (Nov 2011)

Using grid mats with several earth electrodes and using grounding resistance.21.Define the terms feeders and service mains? (Nov 2011)

Feeder is a conductor or transmission line which transmits current from the generating stations todifferent distributing substations. Conductors which connect consumer’s premises with thedistributor are called service mains.

22. What is the function of isolators? (Nov 2013)

Isolator is a knife switch designed to open the circuit under no load condition. It is also called as

disconnector.

23. Mention two significance of neutral grounding.(Nov 2013)

The system voltage during the earth fault depends on neutral earthing. Protection against arcinggrounds, unbalanced voltages with respect to earth, protection from lightning.

24. Give any two factors that affect sag in an overhead line. (Nov 2012)

The factors affecting sag in on overhead line are Wind and Ice loading25. What is meant by string chart? (Nov 2011)

The curves of tension and sag vs. temperature is called string chart.

26. What is meant by sag? (Nov 2013)

The difference in level between points of supports and the lowest point on the conductor is calledsag.

27. What is sag template?(May 2014)

There are two types of supports being used. They are straight and angle tower. While the straightrun towers are used for straight runs and normal conditions, the angle towers are used at angles,terminals and other points where a considerable amount of unbalanced pull may be thrown on thesupports. The angle towers are therefore designed to withstand heavy loadings as compared tostandard towers. Sag template is used to locate and design the angle tower in compare withstraight run tower.

27. What is the need of an earthing system?(Nov 2013)

(i) To save human life from danger of electrical shock or death by blowing a fuse. (ii) To providealternative path for the fault current to flow so that it will not endanger the user. (iii) To protect buildings, machinery & appliances under fault conditions i.e. to ensure that all exposedconductive parts do not reach a dangerous potential (iv) To provide safe path to dissipatelightning and short circuit currents. (v) To provide stable platform for operation of sensitiveelectronic equipment i.e. to maintain the voltage at any part of an electrical system at a known

value so as to prevent over current or excessive voltage on the appliances or equipment. (vi) To provide protection against static electricity from friction.

PART B

1.

Explain in detail the methods of neutral grounding systems.Solid Grounding


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When the neutral point of a 3-phase system (e.g. 3-phase generator, 3-phase transformer etc.) isdirectly connected to earth (i.e. soil) through a wire of negligible resistance and reactance, it iscalled solid grounding or effective grounding.

Resistance Grounding

In order to limit the magnitude of earth fault current, it is a common practice toconnect the neutral point of a 3-phase system to earth through a resistor. This is calledresistance grounding.When the neutral point of a 3-phase system (e.g. 3-phase

generator, 3-phase transformer etc.) is connected to earth (i.e. soil) through a resistor,it is called resistance grounding.

Reactance Grounding

In this system, a reactance is inserted between the neutral and ground 26.15. The

purpose of reactance is to limit the earth fault current. By changing the earthing

reactance, the earth fault current can to changed to obtain the conditions similar to

that of solid grounding

Peterson – coil Grounding

When the value of L of arc suppression coil is such that the fault current IF exactly

balances the capacitive current IC, it is called resonant grounding.2. Explain the classification of substation based on service requirement and constructional feature.

Classification of substations:

(1) According to service requirement

(i) transformer substation

(ii) switching substation

(iii) power factor correction substations

(iv) frequency changer substations

(v) converting substations

(vi) industrial substations

(2) According to constructional features:

(i) indoor substation

(ii) outdoor substation

(iii) underground substation

(iv) pole-mounted substation

3. Write short notes on substation equipments. (Nov 2013)

(i) busbar

(ii) Insulators

(iii) Isolating switches

(iv) circuit breaker

(v) power transformer

(vi) instrument transformer

(vii) metering and indicating instruments

4. Draw the circuit arrangement and explain the various elements of the following bus-bar

arrangements. (i) Single bus scheme.ii) Double busbar with bypass insulator scheme. (Nov2011)(Nov 2013)

Single bus schemeSimplest design and is used for power stations and small outdoor stations having relatively lessincoming or outgoing feeders.


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Double busbar with bypass insulator schemeIt has a main bus bar and spare bus bar. It is important during maintenance and breakdown of the

utility

5. An overhead transmission line conductor having parabolic configuration weighs 1.925 kg permetre of length. The area of X-section of the conductor is 2.2 cm2 and the ultimate strength is

8000 kg/cm2.The supports are 600m apart having 15m difference of levels. Calculate the sag

from the taller of the two supports which must be allowed so that the factor of safety shall beAssume that ice load is 1 kg per metre run and there is no wind pressure.

Sag=45.24m6. A transmission line conductor at a river crossing is supported from two towers at a height of 50

and 80m above water level. The horizontal distance between the towers is 300m. If the tension inthe conductor is 2000 kg, find the clearance between the conductor and water at a point midway

between the towers. Weight of conductor per metre=0.844 kg. Derive the formula. (Nov 2011)

=150 -237= -87m, , D=11.85m, clearance =60.25m abovewater level

,

,

, , , b 7. Deduce an approximate expression for sag in overhead lines when i) supports are at equal levels.

ii) Supports are at unequal levels. (Nov 2012, Nov 2013)

When supports are at equal level

(ii) When the supports are at unequal level

8. An overhead line has a span of 160m of stranded copper conductor between level supports. The

sag is 3.96m at -5.5°C with 9.53mm thick in ice coating and wind pressure of 40Kgf/ of projected area. Calculate the temperature at which the sag will remain the same under conditions

of no ice and no wind. The particulars of the conductor are as follows:Size of theconductor=7/3.45mm, Area of cross section= 64.5m, weight of conductor =0.594 Kgf/m,


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Explain in detail about each topic.

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structure of power system

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