Model Answer Important suggestions to examiners - Msbte ...

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

(ISO/IEC-27001-2005 Certified)

SUMMER– 2019 Examinations

Subject Code: 17417 Model Answer of 36

Important suggestions to examiners:

1) The answers should be examined by key words and not as word-to-word as given in the model answer scheme.

2) The model answer and the answer written by candidate may vary but the examiner may try to assess the understanding level of the candidate.

3) The language errors such as grammatical, spelling errors should not be given more importance. (Not applicable for subject English and communication skills)

4) While assessing figures, examiner may give credit for principle components indicated in a figure. The figures drawn by candidate and model answer may vary. The examiner may give credit for any equivalent figure drawn.

5) Credits may be given step wise for numerical problems. In some cases, the assumed constant values may vary and there may be some difference in the candidate’s answers and model answer.

6) In case some questions credit may be given by judgment on part of examiner of relevant answer based on candidate understands.

7) For programming language papers, credit may be given to any other program based on equivalent concept.

Q.1 Attempt any TEN of the following: 20 Marks a) Draw a single line diagram of A.C supply system.

Ans: Single line diagram of an Electric supply system: ( 2 Marks)

OR Equivalent Figure

OR





b) Classify transmission line according to length of the line. Ans: According to Length of Transmission line: ( 2 Mark)

a) Short Distance Transmission Line - (up to 50 KM)

b) Medium Distance Transmission Line - (up to 50 to 150 KM)

c) Long Distance Transmission Line - (above 150 KM)

OR

1) Short Transmission Line: - The length of Short transmission Line is up to 50KM and

its line voltage is less than 20 KV

2) Medium Transmission Line: - The length of Medium transmission Line is up to

50KM- 150KM and its line voltage is between 20KV to 100 KV

3) Long Transmission Line: - The length of Long transmission Line is above 150KM and

its line voltage is above 100KV

OR

1) Short Transmission Line: - The length of Short transmission Line is up to 80KM and its

line voltage is less than 20 KV

2) Medium Transmission Line: - The length of Medium transmission Line is up to 80KM-

200KM and its line voltage is between 20KV to 100 KV

3) Long Transmission Line: - The length of Long transmission Line is above 200KM and

its line voltage is above 100KV

c) State long form of: (i) AAC (ii) ACSR Ans: Long form of : ( Each long form 1 Mark : Total 2 Marks)

(i) AAC : All Aluminum Conductor

(ii) ACSR : Aluminum conductor steel Re-in force





d) State four properties of conducting materials. Ans: ( Any Four properties expected :1/2 Mark each, Total 2 Marks)

Following are requirements of conductor:-

1. High conductivity :-

Material should have high conductivity

2. High mechanical strength:- Material should have sufficiently high mechanical strength

3. Flexibility:- Material should be flexible

4. Weight:- Material should be light in weight.

5. High resistance to corrosion:- Material should have high resistance to corrosion

6. Brittleness:-

Material should not be brittle.

7. Temperature coefficient of resistance:-

Material should have low temperature coefficient of resistance.

8. Availability & cost:-

Material should be easily available & less costly.

9. Scrap Value:-

Material should have high scrap value.

e) Define regulation of transmission line and write formula. Ans: (Definition 1 Mark & formula 1 Mark ,Total 2 Marks)

Regulation:

Voltage regulation is nothing but voltage drop in transmission line expressed in % of receiving end voltage

% Regulation = 100Re

Re

VoltageEndceiving

VoltageEndceivingVoltageEndSending

f) State any two applications of HVDC transmission system.

Ans: Following are the different applications of HVDC transmission system:-

( Any two Application are expected: 1 Mark each, Total : 2 Mark)

1) HVDC is economical to transmit bulk amount of power (1000 MW) & above Over a





long distance 800 Km & above

2) HVDC is preferred for underground cable transmission as incoming line in megacities.

3) HVDC is preferred for underground cable transmission for crossing long lake, ocean

etc.

4) HVDC is preferred for underground cable transmission where atmospheric conditions

are too bad for overhead transmission line, e.g. High wind pressure, rainfall etc.

5) Interconnection of two transmission lines having different frequencies is possible

through HVDC link.

g) State the function of following layer in construction of a cable: (i) Armouring (ii) Metallic sheathing

Ans: (i) Armouring: ( 1 Mark)

Its purpose is to protect the cable from mechanical injury, while rough handling & at the

time of maintenance.

(ii) Metallic sheathing: ( 1 Mark)

It is provided over insulation to provide the protection of core from entry of moisture,

gases or other damaging liquids (acids & alkaline) in the soil & atmospheric,

h) Why radical system is used for short distances? Ans: Because of following disadvantages radial distribution system is not used for long

distance: (Only two points are expected) ( 2 Marks)

Since there is only one feeder to DTC feed at one point so, 1) There is no reliability to maintain supply at the time of fault on incoming feeder. 2) There is no reliability to maintain supply at the time of maintenance of incoming

feeder. 3) If the system is used for long distance then it takes more time for fault finding &

repairing Hence radial distribution system is used for short distance.

i) What is meant by Ferranti effect of transmission line conductor? Ans: Ferranti effect : ( 2 Marks)

When long distance transmission is lightly loaded or there is no load condition

than it is observe that receiving end voltage (VR) is found to be greater than

sending end voltage (VS). This phenomenon is known as Ferranti effect.





j) State the application of suspension types and pin type insulators. Ans: 1. Application of Suspension type : (1 Mark)

Here one disc is designed for 11 KV so by connecting number of discs in a string of

suspension insulators it can be used for 33/66/110/132/220/400/765 KV & even for

more voltages.

When Line is running straight.

2. Application of Pin type: (1 Mark)

Used for voltage level up to 33 KV maximum.

When Line is running straight.

k) State why three phase four wire system is preferred for secondary distribution system.

Ans: Reason of three phase four wire system is preferred for secondary distribution

system ( 2 Marks)

To obtain 230 volt for single phase supply consumers and to obtain 400/440 volt

for three phase consumers three phase four wire system is preferred for secondary

distribution system.

l) State any two application of HVDC transmission system. Ans: Following are the different applications of HVDC transmission system:-

( Any two Application are expected: 1 Mark each, Total : 2 Mark)

1) HVDC is economical to transmit bulk amount of power (1000 MW) & above. Over a

long distance (800 Km & above)

2) HVDC is preferred for underground cable transmission as incoming line in megacities.

3) HVDC is preferred for underground cable transmission for crossing long lake, ocean

etc.

4) HVDC is preferred for underground cable transmission where atmospheric conditions

are too bad for overhead transmission line, e.g. High wind pressure, rainfall etc.

5) Interconnection of two transmission lines having different frequencies is possible

through HVDC link.





Q.2 Attempt any FOUR of the following: 16 Marks a) Compare overhead line and underground cable (any four points).

Ans: (Any Four points expected: 1 Mark each, Total 4 Marks)

S.No Points Overhead line Underground cable

1 Capital cost Less More 2 Erecting cost Less More 3 Time require for

completion of work Less More

4 Flexibility More flexibility No flexibility 5 Future expansion in

voltage level System voltage can be increased easily

System voltage cannot be increased

6 Overload capacity More Less 7 Fault finding Easy Difficult 8 Charging Current Less More 9 Chances of fault More Less 10 Chances of accident More No chances of accident 11 Safety Less More 12 Radio interference Produces radio interferences Not produces radio

interferences 13 Short cute route Difficult Possible 14 Theft Of energy More possibility Less possibility 15 Voltage drop More Less 16 Power factor Less More 17 Reliability Less More 18 Life Less More

19 Space consumed Space consumed No space consumed 20 Appearance Not good Very good

b) With the help of neat diagram, explain the concept of transposition of conductors. Ans: The concept of transposition of conductors:

( Concept of Transposition: 2 & Figure : 2 Mark , Total 4 Marks)

Transposition of conductor means exchanging the position of 3 phases (R-Y-B) at regular interval.

Each phase occupies 3 different positions consequently on line support (Tower) as shown in fig.

OR Transposition of line conductors means changing the positions of 3 phases on the line





supports twice over the total length of the line

Figure of transposition of conductor:


c) State any four applications where HVDC transmission is used through cable only and not by overhead line.

Ans: Application of High Voltage DC (HVDC) transmission is used through cable:

( Any Four Application Expected: 1 each Point : Total 4 Marks)

1) HVDC is preferred for underground cable when power transmission through

underground cable is greater than 40-50 KM than only HVDC uniquely

suited.

2) HVDC is preferred for underground cable transmission as incoming line in

megacities. . / City centre in- feed.

3) HVDC is preferred for underground cable transmission for crossing long

lake, ocean etc.

4) HVDC is preferred for underground cable transmission where atmospheric

conditions are too bad for overhead transmission line, e.g. High wind

pressure, rainfall, icefall etc.

5) HVDC is preferred for underground cable for long distance underwater

power links.

6) HVDC is preferred for underground cable for powering island from onshore.

7) HVDC is preferred for underground cable for taking power from offshore

wind farm.





8) HVDC is preferred for underground cable for powering oil and gas offshore

floating platform.

d) State the requirements of a distribution system. Ans: Following requirements of a distribution system.

( Only Four requirement expected 1 Mark each : Total : 4 Marks)

1. Layout should be simple in design.

2. It should have less initial cost

3. The distribution system should have minimum distribution losses.

4. Voltage drop in distribution system should be less and within permissible limit ( 6%).

5. From safety point of view distribution system should maintain proper clearances.

6. The rating of distribution transformer & cross section of conductor should be

proportional to result of load densities present & future.

7. Power should be available to consumers whenever needed.

8. A steady, non-fluctuating, quality supply (Pure sine wave) should be available to

consumers.

9. Distribution system should not be over loaded.

10. Distribution system should have high reliability to maintain supply.

11. Distribution system lay out should not affect the appearance of locality.

12. Before installation of distribution system proposed widening of the road in the

near future are to be kept in mind

13. It should have low, easy, less costly & less time consuming maintenance.

14. Fault on nearest distribution system should not affect stability of existing

distribution system.

15. Time required for completion of work should be less.





e) A three phase overhead transmission line is being supported by 3 disc insulators. The potentials across top unit and middle units are 8 kV and 11 kV respectively. Calculate: (i) Line voltage (ii) String efficiency

Ans: Given Data: V1= 8KV, V2= 11KV i) Ratio of capacitance ‘k’ :- V2 = V1 (1+ m) 11 = 8 (1+ 8m) 11 = 8 + 8m 3 = 8m k = m = 0.375 ------------------------------------ (1/2 Mark)

V3 = V1 (m2 + 3m + 1)

= 8 ( 0.3752 + (3x 0.375) +1)

V3 = 18.125 KV ---------------------------------------- (1 Mark)

Voltage across string = Vph = V1 +V2+ V3 = 8+11+18.125 = 37.125 KV ------------------------------- (1/2 Mark)

ii) The line voltage VL = phV3

VL = 125.373

VL = 64.30KV ---------------------------------------- (1 Mark) iii) String efficiency:-

String 1003

00

V

Vph

String 100125.183

125.370

0

String 00 = 68.27 %. ---------------------------------------- (1 Mark)

f) Compare outdoor and indoor substation. Ans: ( Any four point expected: 1 Mark each point: Total: 4 Marks)

S.No Points Outdoor Substation Indoor Substation

1 Capital cost Less, as construction work cost is less.

High, as construction work cost is more.

2 Time required Less, as construction work is More, as construction work is





for completion less. more.

3 Distance between two equipment

More, this will reduce possibility of fault & safety increases

Less, this will increase possibility of fault & safety reduces.

4 Access for incoming & outgoing line

Easy access for incoming & outgoing lines because of outdoor installation.

Difficult access for incoming & outgoing lines because of indoor installation.

5 Cooling arrangement

Natural cooling is available due to outdoor installation. This reduces energy consumption charges due to outdoor installation.

Natural cooling is not available so artificial cooling arrangement is required This increases energy consumption charges due to indoor installation.

6 Availability of natural light

Natural light is available in day time, so there is no need of illumination during day time. So it saves electrical energy & its cost

Natural light is not available even in day time, so there is need of illumination even during a day time. This increases energy consumption charges due to indoor installation

7 Detection of fault

Easy, as all equipment’s are easily viewed.

Difficult, as all equipment’s are not easily viewed.

8 Replacement of equipment

Easy, due to outdoor installation. Difficult, due to indoor installation.

9 Future expansion

Expansion of substation is easily possible whenever needed & can be completed in less time & cost.

Expansion of substation is not easily possible whenever needed because of construction work. Also it require more time & cost.

10 In case of accident

In case of accident there is less risk & damage to other equipment’s than indoor substation.

In case of accident there is more risk & damage to other equipment’s than outdoor substation.

11 Space Require More Less





12 Effect of atmospheric condition

Switching operation is difficult in rainy season & it is less safe

Switching operation is not difficult in rainy season & it is more safe due to indoor installation

13 Chances of leakage current

More due to outdoor installation Less due to indoor installation

14 Maintenance cost

More due to outdoor installation. Less due to indoor installation

15 Applications Where atmospheric conditions are clean and dry also where space available is more than subs stations are installed outdoor.

In places where heavy rainfall, snow fall occurs or there is humidity in atmosphere also where availability of space is less then under such situations sub stations are installed indoor.

Q.3 Attempt any FOUR of the following: 16 Marks a) Under which conditions Ferranti effect occurs. State any four conditions? What is

Ferranti effect? Ans: Under Following conditions Ferranti effect occurs:

( Any Four condition expected: 1/2 Mark each : Total : 2 Marks)

1. When there is no load on transmission line (IL = 0) OR

2. When There is no load at receiving sub-station or Lightly loaded OR

3. When there is sudden load thrown OFF. OR

4. When there is sudden load shading. OR

5. When Transmission line is open circuited due to load failure.

Ferranti effect : ( 2 Marks)

Under any one above condition, it is oberve that receiving end voltage (VR) is

found to be greater than sending end voltage (VS). This phenomenon is known as

Ferranti effect.





b) State any four factors which affects corona. State two points how corona effect can be reduced.

Ans: The Following Factors affecting corona:- ( Any Four Factors expected: 1/2 Mark each : Total : 2 Marks)

1. Magnitude of Voltage :

If voltage across two conductors is greater than 30 KV/cm, i.e. breakdown

voltage of air than corona formation starts. Corona will not start if voltage is below

30 KV/cm

2. Distance between two conductor:

If spacing between two conductors is very large as compare to their diameter

than there is no possibility of corona formation. Because value of voltage at which

corona occurs increases.

3. Size of conductor:

If size (Cross section) of conductor is more, than magnitude of voltage

required to occur the corona increases.

4. Condition of conductor & Hardware:

Rough and irregular surface of conductor and hardware will give more

corona than solid, smooth body conductor & hardware.

5. Atmospheric Condition:

As corona takes place due to ionization of air so it depends on condition of

air so for dry air formation of corona occurs late than in wet air (damp atmosphere

condition/ rainy season/thunderstorms/fog air becomes more conductivity)

6. Effect of supply Frequency:

Corona loss varies directly as the supply frequency

7. Effect of density of air:

Corona loss increases with the decrease in the density of air (The

corona loss of transmission line passing through hilly area is higher than that of a

similar line in plain due to reduced value of air density at high level /altitude)





As Following points Corona effect can be reduced:

( Any Two Point expected: 1 Mark each : Total : 2 Marks)

1. By increasing distance between two conductor i.e. by using longer cross arm.

2. By using larger size(diameter) of conductor e.g./ using ACSR, bundled conductor

3. By using smooth body conductor and hardware.

c) State generalized constants A, B, C, D of formula for nominal `T' network. Ans: Generalized constants A, B, C, D of formula for nominal `T' network: ( 4 Marks)

phR

phS

V

VA ------------ Voltage Ratio

R

phS

I

VB ------------ Impedance in ohm

phR

S

V

IC ------------ Conductance in mho

R

S

I

ID ------------ Current Ratio

A, B, C, D constants are generally complex number

Values of ABCD constants T-equivalent circuits of are as bellows:

A = D =2

1YZ

B = Z 4

1YZ

ohm

C = Y mho





d) Draw a neat layout of Grid distribution scheme. State two advantages of it.

Ans:

Layout of Grid distribution scheme: ( 2 Marks)

OR


Advantages of Grid distribution scheme:

( Any Two point expected: 1 Mark each : Total 2 Marks)

1. Supply to distribution transformer center is given through two different





generating stations or major generating stations

2. It has highest reliability to maintain supply even when there is a fault on any

one feeder

3. It has highest reliability to maintain supply even when there was maintenance

on any one feeder.

e) Draw a typical layout diagram of 11 kV distribution substation. Ans: Typical layout diagram of I1 kV distribution substation ( 4 Mark )

OR

OR





f) A single phase 11 kV with a length of 15 km is to transmit 500 kVA. The inductive reactance of the line is 0.5 ohm/km and resistance is 0.3 ohm/km. Calculate the efficiency and regulation of the line for 0.8 lagging p.f. Draw vector diagram.

Ans: Length of line =15km, PR = 500KVA, VR = 11KV, P.F. = 0.8 lag, R Per conductor =0.3ohm/km, X Per conductor = 0.5

ohm /km

Step 1: To calculate current

Power in KW= KVAP.F.

= 5000.8 = 400Kw

Power P = cosVI

cosV

PI

, 8.011

400

I

ampI 45.45 ------------------------------------- ( 1/2 Marks)

Step 2: To calculate value of sin :

6.0sin8.0 RRCos

R=0.3 ohm/km x 15= 4.5 ohm

X= 0.5ohm/km x 15= 7.5 ohm





To Calculate Total /loop values of R & X

Total resistance RT = 2R= 2 x 4.5 = 9 ohm

Total Reactance XT = 2 X =27.5 =15 ohm

Step 3: To calculate Sending end voltage:

Vs = RTRTR XRIV sincos

= 6.0158.0945.451011 3

= 11736.29 volt

Vs = 11.73629 KV ---------------------------------------- ( 1/2 Marks)

Step 4: To calculate Total Line Losses:

Total Line Losses = TRI 2

= 9)45.45( 2

= 18591.323 Watt

= 18.591323 KW ------------------------------------- ( 1/2 Marks)

Step 5:To calculate voltage regulation:

% Voltage Regulation = 100

R

RS

V

VV

= 10011

1173629.11

% voltage regulation: = 6.69 % ----------------------------------- ( 1 Marks)

Step 6: To calculate Total Transmission efficiency:

100200

TR

RT RIP

P

100323.1859110400

104003

3

00

T

00

00 56.95T ---------------------------- ( 1 Marks)





Step 7: Vector Diagram: ------------------------------------- ( 1/2 Marks)

Q.4 Attempt any FOUR of the following: 16 Marks a) Draw a neat labeled diagram of an underground cable to show different parts.

Ans: Neat diagram show the various parts of an underground core cable. ( 4 Marks)

OR





b) What is skin effect? On which factors does skin effect depend? Ans: ( Meaning : 2 Marks & Factors: 2 Marks, Total 4 Marks)

Skin effect:- (2 Marks)

When alternating current flows through conductor it has tendency to flow away from

center of conductor.

i.e. maximum current density is near skin of conductor and goes on reducing towards

center core is known as skin effect.( Since the inductive reactance (XL) at the center of the

conductor is more than surface of conductor)

OR

The tendency of alternating current to concentrate near the surface of a conductor

is known as skin effect.

Skin effect depends on following factors: (2 Marks)

1. Supply frequency: As frequency increases skin effect increases.

2. Cross section of conductor: Skin effect increases with increase in diameter of conductor.

3. Solid conductors have more skin effect than stranded conductors.

4. Permeability of conductor material.

c) A single phase overhead transmission line draws 1100 kW at 33 kV 0.8 p.f. logging. The total resistance and inductive reactance at the line are 10 ohm and 15 ohm respectively. Determine the sending end voltage and sending end p.f.

Ans: Given Data:-

PR = 1100KW VR = 33KV P.F. = 0.8 lag RT = 10 ohm XT = 15 ohm

Step 1: To calculate current:

Power P = cosVI

cosV

PI

, 8.033

1100

I

ampI 67.41 --------------------------------- ( 1 Marks)

Step 2: To calculate value of sin :

6.0sin;8.0 RRCos ------------------------------------- ( 1 Marks)






Vs = RTRTR XRIV sincos

= 6.0158.01067.411033 3

= 33708.39volt

Vs = 33.70839 KV ------------------------------------- ( 1 Marks)

Step 4: To calculate sending end Power factor:

S

TRR

V

IRVs

coscos

3

3

1070839.33

1067.418.01033cos

s

lags 7955.0cos ------------------------------------- ( 1 Marks)

d)

Study the Fig. No. 1 and answer following questions:

(i) Which part is shown by 'A'? (ii) State the meaning of symbol shown at 'B' point. (iii) State the voltage rating of equipment at point `D'. (iv) Which part is shown by 'E'?

Ans: ( Each answer : 1 Mark, Total 4 Marks)

(i) Part is shown by 'A' : Generation

(ii) The meaning of symbol shown at 'B' point : Step-up transformer Substation

(iii) The voltage rating of equipment at point `D'.: Distribution transformer Substation

(iv) Which part is shown by 'E' : Primary transmission line ( 3-ph, 3 wire)





e) State and explain any one method for improving string efficiency. Ans: The Methods of Improving String Efficiency:-

( Methods : 2 Mark & Any one explanation: 2 Marks : Total 4 Marks)

1) By reducing value of ‘m’ or (‘k’) by using longer cross arm.

2) By Making of ‘m’ or (‘k’) equal to zero

3) By grading of Insulator.

4) By Using guard ring.

Explanation:-

1) By reducing value of ‘m’ or (‘k’) by using longer cross arm:-

Fig:-

The value of ‘m’ can be decreased by reducing value of shunt capacitance (C1) since m = C1/C. In order to reduce value shunt capacitance (C1) distance of string of insulator from tower must be increased. i.e. by using longer cross arm. Due to this value of shunt capacitance (C1) reduces.

Therefore value of m reduces Since (C

Cm 1 ) As value of ‘m’ reduces

there will be more uniform voltage distribution along a string of suspension insulator. In this way string efficiency increases. Limitation:

In practice there is limitation to increase length of cross arm as cost of tower increases. In practice m= 0.1 is the limit which can be achieved by this method.

2) By Making of ‘m’ or (‘k’) equal to zero:-

or equivalent Figure





If an insulating material or any non conducting material of high strength is

used for connection between two disc insulators in a string instead of using steel

part.

Than value of Shunt Capacitance (C1) becomes Zero,(Capacitance will not

form) therefore value of ‘m’ becomes zero (since m = C1/C) So string efficiency

becomes 100%.

3) By grading Insulator :-

In this method, disc insulators of different dimensions are so selected that each

disc has different capacitance. The assembly in the string of suspension insulator is

made in such a way that the top unit insulator has fewer dimensions. (Less

capacitance) (C A) and dimensions of insulators progressively goes on increasing

i.e. bottom unit has maximum capacitance due to large dimensions of insulators.

(Since Q=C/V i.e. V is inversely proportional to capacitance So as A Increases C

increases therefore voltage decreases)

In this way it equalizer potential distribution across the string and therefore

increase string efficiency.

This method has disadvantages that it requires disc insulator of different

dimensions in one string of suspension insulator. Practically it is not possible to

obtain such ration. But very high voltage transmission line (1200KV). This method is

used.

4) By Using guard ring :-

or equivalent Figure





Guard ring is a metal ring electrically connected to conductor and surrounding

the bottom insulator.

Due to guard ring leakage current through all discs in a string is same.

So, we will get uniform voltage distribution along the string of suspension

insulator, In this way string efficiency increases.

f) Write down the functions of following elements of a substation: (i) CB (ii) Relays (iii) Lightning arrester (iv) Isolators

Ans: (i) CB ( Circuit Breaker) ( 1 Mark)

It is protective device. It open or break the circuit whenever there is fault &

protect the equipment. It can be operated manually or remote control whenever

required.

ii) Relay: ( 1 Mark)

It sense the faults & gives signal to trip circuit of C.B. to open. There are

different types of relay e.g. Earth fault relay, Phase to Phase/short circuit fault

relay, Thermal relay etc.

iii) Lightning Arrester: - ( 1 Mark)

It is provided for protection of substation, transformer against lightning

stroke .It is connected in between line and ground at the starting point of substation.

Under normal condition it acts as an insulator.

iv) Isolator : - ( 1 Mark)

Its function is to connect or disconnect the circuit only when there is no load.

Isolator is inter locked with C.B. Isolator can be operated by hand locally as well as

by motorized mechanism from remote position. Normally it is closed.





Q.5 Attempt any FOUR of the following: 16 Marks

a) State the effect of lag, lead and unity power factor on regulation of transmission line with phasor diagram.

Ans: (Vector diagram 3 Mark, Each effect on regulation 1 Mark , Total 4 Marks)

i) Vector Diagram for Lagging Power Factor:-

or equivalent figure

EFFECT :- At Lagging PF Receiving voltage is less than Sending end hence regulation

is positive

ii) Vector Diagram for Leading Power Factor:-


EFFECT: - At Leading PF Receiving voltage is more than Sending end hence

regulation is negative.

iii) Vector Diagram for Unity Power Factor:


EFFECT :- At UPF receiving voltage is less than Sending end hence regulation is positive





b)

A single phase AC distributors of 900 m length has total impedance of (0.02 + 0.04j) ohm and is fed from one end at 250 V. If it is loaded as in Fig. No 2. Calculate the voltage drop and voltage at far end.

Ans:

04.002.0 jZ

givenimpedanceTotalZT 43.630447.0

Step 1: To find section impedance:

04.002.0900

300jZAB

)04.002.0(33.0 ABZ

)0132.00066.0 JZ AB

434.630147.00 ABZ

434.630147.00 CDBCAB ZZZ

Step 2 : To Calculate section current :

lagfpatID 7.0..,60

AI D 5729.4560

AjID 8485.4242





lagfpatAIC 8.0..,60

86.3660 CI

AjIC 3600.48

ALfpUnityatAIB0060)(..,60

AjI B 060 ----------------------------------------------------------------(1/2 Marks)

DCBC III

)8485.4242()3600.48( jjI BC

8485.7890 jIBC

AIBC 2214.416540.119

DCD II

AjICD 8485.4242

DCBAB IIII

)8485.4242()3600.48()060( jjjI AB

8485.78150 jI AB

AI AB 72.2746.169 -- ------------------------------------------(1/2 Marks)

Step-3: To Calculate Section voltage drop

04.002.0 jZ

givenimpedanceTotalZT 43.630447.0

CDBCAB ZZZ

434.630147.00 CDBCAB ZZZ --------------------------------- (1/2 Marks)

Voltage drop in section CD:-

= ICD X ZCD

= )434.630147.0()5729.4560( 00





= Volts0864.17882.0

VCD = Voltsj 27.0839.0 ------------------------------------ (1/2 Marks)

Voltage drop in section BC:-

= IBC X ZBC

= )434.630147.0()22.416540.119(

= Volts0214.2276.1

VBC = Voltsj 6654.0629.1 ----------------------------------- (1/2 Marks)

Voltage drop in section AB:-

= IAB X ZAB

= )434.630147.0()72.2746.1169(

= Volts714.3549.2

VAB = Voltsj 454.102.2 ------------------------------------(1/2 Marks)

Total Voltage drop from A to D:-

= VAB+ VBC +VCD

= )27.0839.0()6654.0629.1()454.1021.2( jjj

= Voltsj 3894.2489.4

= Volts00254.280853.5 ---------------------------------- (1/2 Marks)

Voltage at far end:- DtoAfromdropvolatgeTotalvolatgeendSending

= )0250( j )3894.2489.4( j

= Voltsj 3842.2511.245

= Volts05576.05226.245 --------------------------------- (1/2 Marks)

c) Write down any four properties of line support. Ans: Following are requirements or properties of the line supports

( Any Four Point expected: 1 Mark each point, Total 4 Marks)

1. High mechanical strength:-

It should have high mechanical strength to withstand against -

Wind pressure





Load of fabrication

Weight of Insulator

Weight of conductor etc.

2. Light in weight:-

It should be light in weight to reduce-

Transportation cost

Handling, loading, unloading cost and

Erection cost.

3. Effect of atmospheric conditions: It should be withstand even at bad

atmospheric condition.

4. High resistance to corrosion: It should have high resistance to corrosion to avoid

rusting.

5. Initial & Maintenance cost: It should be less.

6. Easy access: It should be easily accessible for wireman for line work and

maintenance work. or They must be easily accessible for point and erection of

line conductors

7. Life: It should have longer life.

8. Appearance: It should have good appearance or They must be of pleasing shape

d) Draw nominal ` network. Draw its phasor diagram

Ans: ( Diagram: 2 Mark & Phasor diagram: 2 Mark : Total : 4 Marks)

Circuit Diagram Nominal (pi) Network: -

or equivalent diagram





Vector diagram of network for:

OR

e) State difference between feeder and distributor on any four points.

Ans: ( Any Four points are Expected: 1 each Point : Total 4 Marks)

Sr.No Feeder Distributor

1 It is link between receiving substation & distribution transformer

It is link between distribution transformer substation & consumer

2 It is also called as a High Tension Line

It is also called as a low Tension Line

3 It is a 3-Ph, 3 wire system.(R-Y-B) It is a 3-Ph, 4 wires system. (R-Y-B-N)

4 Feeder voltage is 11KV/22KV/33KV depending upon load

Distributor voltage is for 3-ph consumer- 400V and 1-Ph consumer- 230V

5 Feeder is high capacity conductors. Distributors are low capacity





conductors

6 Feeder forms the primary distribution system

Distributors forms secondary distributor system.

7 While designing feeder its current carrying capacity is important

While designing distributor its voltage drop calculation is important.

8 Feeder is not tapped along its length

Distributors are tapped throughout its length.

9 Its loading point is at substation only

Distributors loading point is throughout its length.

f) State the factors to be considered while deciding location of site for substations. Ans: Following factors should be considered while deciding location of site for sub-station:-

( Any four point expected : 1 Mark each, Total 4 Mark )

1. Near load center:

Sub-station should be located near load centre to reduce cost of Transmission and

distribution lines and to reduce losses in it.

2. Easy access for transmission Line :

There should be easy access for incoming and outgoing line.

3. Easy access towards sub-station :-

There should be easy access towards sub-station for transportation of equipments and

manpower etc.

4. Space(Land ) available :

The land proposed for a substation should be normally level and open from all sides &

sufficient land should be available for installation of sub-station and future expansion. e.g.

5. Atmospheric conditions :

Atmospheric condition in the area of sub-station should be clean and dry also There should

be less atmospheric pollution.

5. Cost of land :





Cost of land should be less to reduce capital cost of sub-station.

6. Municipal restriction :

Where municipal restriction will not take any objection for required type building of sub-

station.

7. Staff amenities :

The site should be such that essential amenities must be available to staff like residential

quarters, drinking water, school, hospital, public transportation, communication.

8. Bearing capacity of land (Hard land ):

To reduce construction cost of building and for better foundation of equipments land

should have high bearing capacity.

9. Area free from earthquake :

To avoid damage to sub-station area should be free earth quake.

Q.6 Attempt any FOUR of the following: 16 Marks a) Draw a neat sketch of double circuit RCC pole.

Ans: Neat sketch of double circuit RCC Pole: ( 4 Marks)






b) Draw diagram of end condenser method. Also draw phasor diagram. Ans: Diagram of end condenser method:

(Diagram : 2 Mark & Phasor diagram: 2 Mark, Total : 4 Mark)


Phasor diagram of end condenser method:


c) Compare HVDC transmission with EHVAC transmission. Ans: (Any four Point Expected : 1 Mark each, Total 4 Marks)

S.N

o

Points HVDC EHVAC

1 Number of conductor required for single circuit

One conductor.& Ground is used as a return path

Three conductors (R.Y.B)

2 For double circuit Two conductors.& Ground is used as a return path

Six conductors (R,Y,B & R,Y,B)

3 Design of Tower Light Heavy

4 Intermediate substation Not required Required at every 250

Km





5 Capital cost of S/S More Less

6 Transmission line cost/km for long distance. (Above 500km.)

Less More

7 Ground return Possible Not possible

8 Frequency Absent Present

9 Skin effect Absent Present

10 Proximity effect Absent Present

11 Ferranti effect Absent Present

12 Corona losses Less More

13 Radio interference Absent Present

14 Effect of L &C Absent Present

15 Value of resistance Less More 1.6 times than DC

16 Copper loss Less More

17 Transmission Efficiency More Less

18 Voltage drop in transmission line

Less More

19 % Regulation Better Good 20 Limitation on length of

cable Charging current is absent so no limitation on length of cable

Due to charging current there is limitation on length of cable

21 String efficiency 100 % Less than 100 %

22 Losses in S/s More Less

23 Maintenance cost of S/S More Less

24 Asynchronous tie Possible Not possible

25 Reliability & availability One bipolar line is sufficient

AC Double circuit are necessary

26 Control system Difficult, costly Simpler cheaper 27 Power handling capacity No limit There is limit due to

inductance & power





angle 28 Voltage control for long

distance lines Easier as L&C are not effective

Difficult for long distance lines due to presence of L &C

29 Stability limit No limit due to absent

of inductance &

power angle

EHVAC limits due to

inductance & power

angle

30 Power flow control Power can be quickly(fast) controlled,

Power flow cannot be easily controlled, (slow)

31 Power transfer ability High Lower

32 Transient performance Excellent Poor

33 Back to Back conversion stations

Possible Not Possible

34 Short-circuit current

level

Less More

35 Reliable circuit breaker Not available Available

36 Fault levels Remains unchanged Get added after

interconnection

37 Frequency conversion Possible Not possible

38 Cascade tripping of

circuit

Avoided Likely

39 Spinning reserve Reduced Not much reduced

40 Frequency of fault Less More

d) Draw ring main system and give its advantages. Ans: Ring main system of distribution with neat diagram:

( Figure ; 2 Mark & Advantages: 2 Marks)






Advantages:-(Any TWO point is expected)

1. Supply to distribution transformer center is given through two different Feeders

2. Reliability to maintain supply is more even when there is a fault on any one feeder.

3. Reliability to maintain supply is more even when there was maintenance on any

one feeder

e) Describe the effect of load p.f on performance of transmission line.

Ans: Effect of load p.f on performance of transmission line:

1. Effect of improved power factor on Efficiency:- ( 2 Mark) As power factor increases, current decreases, so Copper losses decreases, Hence

transmission efficiency increases & vice versa.

2. Effect of improved power factor on Regulation:- ( 2 Mark)

As power factor increases, current decreases, So Voltage drop in transmission line

decreases, As a result, regulation get improved (decrease) an vice versa.

OR

1. Effect of poor power factor on efficiency:- ( 2 Mark)





When power factor of load reduces current drawn by transmission line increases so copper losses in transmission line increases, hence transmission efficiency reduces.

2. Effect of poor power factor on voltage Regulation:- ( 2 Mark) When power factor of load reduces current through transmission line

increases, so voltage drop in transmission line (due to resistance & inductive reactance) increases so regulation increases. (Become Poor)

f)

A single line 11 kV short transmission line delivers 1000 kW power at 0.8 p.f. lagging total resistance and inductive resistance and inductive reactance of the line are 5 ohm and 5.6 ohm. Determine: (1) Sending end voltage.(2) Percentage regulation of transmission line.

Ans: 0.8 sin 0.6R RCos Resistance 5.0 ohm Reactance 5.6 ohm Step 1: To calculate Sending end voltage: Power P = cosVI .------------------------------------------- (1/2 Mark)

cosV

PI

, 8.01011

1010003

3

I

ampI 6363.113 .--------------------------------------------- (1/2 Mark)


Vs = RTRTR XRIV sincos .------------------------ (1 Mark)

= 6.06.58.056363.1131011 3

= 11000 +72.7272

= 11072.7272 volt

Vs = 11.07272 KV .----------------------------------- (1/2 Mark)

Step 4:To calculate voltage regulation:

% Voltage Regulation = 100

R

RS

V

VV.------------------------ (1 Mark)

= 10011000

110007272.11072

= 0.6611 % ----------------------------------- ( 1/2 Mark)

------------------------------------------------------END-------------------------------------------------------

Model Answer Important suggestions to examiners - Msbte ...

Documents

Transcript of Model Answer Important suggestions to examiners - Msbte ...