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D1

D2 D4

D3CH1

CH2

CH3

CH4

I0

V0+

La E

+

i

VS

01. (a)

Sol:

From the above diagram choppers shown are

switches these are self commutated by the

feed back diodes shown as D1, D3, D4, D2.

For duty ratio greater than 0.5 voltage is

‘ve’ and for < 0.5, voltage is ‘+ve’

Here operated means corresponding switch

is get ON and OFF in that period

When < 0.5

First quadrant operation

CH4 and CH1 is ON: so current flows

through Vs – CH1 – La – E – CH4 - VS

So V0 is +ve, I0 is +ve

When CH1 is turned off, positive current

free wheels through CH4, D2. In this manner

both V0, I0 can be controlled.

Second Quadrant operation

Here CH2 operated and CH1, CH3 and CH4

are kept off, with CH2 ON, reverse (or

negative) current flows through La, CH2, D4

and E. Inductance La stores energy during

the time CH2 ON.

In this time La stores energy then the load

voltage (V0 = E +dt

diL ) > VS

When CH4 is turned off ‘V0’ will turn on

diode D1, D4 then the current is fed back to

source through D1, D4 as load ‘V0 is +ve and

I0 is negative.

When > 0.5 (V0 is negative)

Third Quadarant operation: (V0 is ve, I0

is ve). CH1 is kept off, CH2 is kept ON and

CH3 is operated. Polarity of load emf ‘E’

must be reversed for this Quadarant working

V0 = E0 + I0R

When CH3 ON, load gets connected to

source ‘VS’ so V0 I0 are negative leading to

third Quadarant operation. When CH3 is

turned off. load voltage turn on diode D4.

Load current free wheels through CH2, D4

Fourth Quadrant operation: (V0 ’Ve’, I0

‘+ ve’)

Here CH4 is operated and other devices kept

off. Load emf ‘E’ must have its polarity

reversed. When CH4 ON positive current

flows through CH4, D2, La and E.

CH4 operatedCH4-D2 : La stores energyCH4:OFF: then D2-D3

conduct

CH2 operatedCH2-D4: La stores energyCH2 OFF: then D1-D4

conduct

CH1-operatedCH1, CH4 ON`CH1 OFF: then CH4-D2

Conduct`

CH3 operatedCH3-CH2 ONCH3, OFF: then CH2 – D4

conductV0

V0

I0I0

: 3 : ESE MAINS


Inductance La stores energy during this

period.

This energy will turn on diodes D2, D3 when

CH4 turned off. So that current is fed back

to source through D2, D3. Also power is fed

back from load to source.

01. (b)

Sol: Pr = sin.|B|

|V||V| rs

220220

170150sin

= 0.5269

= 31.79

Now, 2r

rsr |V|

|B|

|A|cos

|B|

|V||V|Q

2o 220170

93.079.31cos

170

220220

= 284.705 0.85 – 246.77

= –4.77 MVAr

222r

2r

rr

77.4150

150

QP

Pcos

0759.150

150

= 0.999 lead

01. (c)

Sol: Operation at rated conditions:

KVL: 250 = 6.4 + E; so E = 243.6 V.

E = KIfm

m =3

10060

)1000(2

r/s

KIf100

3

= 243.6

which gives KIf = 2.3262

The developed torque = K If Ia = 2.3262

160

= 372.2 N-m/r.

Load torque is given to be constant,

irrespective of speed variations.

If torque for friction and core losses is also

assumed constant, the developed torque will

remain constant, irrespective of speed

variations.

(i) When field flux is reduced to 70% of its

original value but we still want Td to be

unchanged, then Ia = (160/0.7) = 228.6 A.

Corresponding emf = 240.86 V.

(ii) 2.32620.7m = 240.86

From which m = 147.9 r/sec or 1412.5 rpm.

160A+

dcsupply

–

E

0.04

+

–

250V6.4V

+

1000 rpm

: 4 : ESE MAINS


: 5 : ESE MAINS


01. (d)

Sol: Z = (5 + j20)

PD = 30 MW, pf = 0.8 lag

V = 33 kV, f = 50 Hz

VS = VR = 33 kV, QC = Z = 20.6276

IR = Is =8.0333

1030 3

= 0.656136.86 kA

(i) PD = PR = 30 MW

PR = 30 = 76cos62.20

)33()76cos(

62.20

)33( 22

PR = cosZ

V)cos(

Z

V 2

R

2

R

PR = ]cos)[cos(Z

V 2

R

From the above equation,

= 40.1

QR = sinZ

Vsin

Z

V 2

R

2

R

= 76sin62.20

)33()1.4076sin(

62.20

)33( 22

= 30.96 51.25

QR = –20.29 MVAR

QC (3-) = Qload QR

= 308.0

6.0 (20.29) = 42.79 MVAR

QC (3-) =42.79 MVAr leading

(ii)

QC/Ph =3

79.42 = 14.26 MVAR

QC/ph =/phX

V

c

2Ph

QC/ph = f2.V2ph C/ph

C/ph =f2.V

ph/Q2ph

C

=

502π(33)

14.262

= 41.6 F

01. (e)

Sol: Given data:

3.5 MVA, 10 kV

3 – ; N = 600 rpm

No. of poles, p = 10

Low speed alternator

No. of slots ,S = 144

Two layer (i.e., Double layer winding)

Conductors / coil side = 5 in each slot

Coil span = 12 slots

Flux/ pole, = 0.116 Wb

Calculate Eph(r.m.s)

∵ Conductor/coil side = 5 in each slot

no. of conductors / slot = 10

Total no. of conductors = 10 × 144

= 1440

33kV

C/phC/ph

C/ph

: 6 : ESE MAINS


Total no. of turns =2

1440 = 720

Turns/ph =3

720 = 240

Slot angle () =S180p

=144

18010 = 12.5

Coil span = 12 slots = 12 × 12.5 = 150

Short pitch angle () = 180 –

= 180 – 150 = 30

Kp = cos2

=2

30cos = 0.965

Slots/ pole/phase

310144

m

= 4.8

(∵ ‘m’ is in fraction so it is fractional wdg)

Kd =

2sinm

2m

sin

=

25.12

sin8.4

25.12

8.4sin = 0.956

f =120

60010120PN = 50 Hz

Eph = 4.44 Kp Kd f Tph

= 4.44×0.965×0.956×0.116×50×240

= 5.7 kV

02 (a) (i)

Sol: When source voltage is dc ,since the series

motors have very high starting torques, they

are best suited for traction drives. The below

figure(a) shows the basic chopper circuit for

the control of a dc series motor. Here also

one can employ fixed frequency TRC (or)

variable frequency TRC.

The voltage and current wave forms shown

in below figure

During ON-time of the chopper, the

supply voltage Vdc , applied to the

motor.

During OFF-period, of the chopper, no

voltage is applied to the motor.

The main problem in the analysis of a

chopper-controlled series motor araises due

to the nonlinear relationship between

armature induced voltage and armature

current because of saturation in the

magnetisizing characteristic.

The analysis of series motor for TRC is now

carried out by making the following

assumptions.

MVdc

CH1

Eb

+

–

V0

field

Figure (a) Chopper control of dc series motor

V0I0

Armature current

Voltage

t

Fig. Voltage and current waveforms

: 7 : ESE MAINS


(i) Thyristor (T) and diode (D) are assumed

to be ideal one.

(ii) Motor armature speed is assumed to be

constant during one complete chopper

cycle.

During on Period:

bfaa

afadc eRRi

dt

di)LLV

ONT

0 bfaaa

faavg dteRRidt

diLL

T

1V

Eb = KNia

During OFF Period:

bfaaa

fa eRRidt

diLL0

Where,

K = Motor constance

N = Speed of motor

Vavg = Iavg (Ra +Rf) +[Eb]avg

Steady state torque Tavg Iavg

= flux in motor field

Tavg =2avg

2avg IKI

offon

ondcavg TT

T.VV

The speed – torque characteristics of series

motor is shown below

02. (a) (ii)

Sol: (a) For zero degree firing angle, motor

terminal voltage is rated i.e. 230V. therefore,

V230V0cosV23

t

Or V34.170123

230V

Here Vl is the line voltage Per-phase voltage

on transformer star side is

V35.983

34.170Vph

Per-phase voltage input to transformer delta

= 400V

Transformer phase turns ratio from

primary to secondary

067.435.98

400

(b) (i)

At 1500 rpm, Ea =Vt – Iara

= 230 – 20 0.6

= 218V

At 1000 rpm, motor emf

V33.14510001500

218

For this motor emf, armature terminal

voltage at rated torque is

Vt = Ea + Iara = 145.33+200.6 = 157.33V

But V33.157VVcosV3

t0m

Torque

Ton2

Ton1

Ton1 > Ton2

Speed

Speed –Torque characteristics

: 8 : ESE MAINS


Or

84.4634.17023

33.157cos 1

(ii) At half the rated torque, armature current

currentrated2

1Ia

= A10202

1

6.0102181500

900cos

34.17023

V8.124

Or

861.12234.17023

8.124cos 1

02. (b)

Sol: String efficiency: It is a measure of the

utilization of material in the string and is

defined as

conductorpowerthenearunittheacrossvoltagen

stringtheacrossvoltage

Let ‘V’ be the operating voltage and V1, V2

are the voltage drops across the units

starting from the cross arm towards the

power conductor.

Arcing Horns:

Arcing horns are used to protect the

insulators from voltage surges.

These are the projecting conductors which

are arranged across the surface of the

insulator to be protected during flashover.

These are generally arranged on paired on

either side of the insulator strings of

overhead lines and bushings of the

transformer, then the length of the string is

increases. The larger the number of insulator

units, the longer will be the string, the

voltage distribution across various insulator

units will be non-uniform. A uniform

distribution of voltage across the insulator

units of a string can be obtained by

providing a large metal ring called the guard

ring connected to the metal work at the

bottom of the unit.

The guard ring, when used a long with the

arcing horn fixed at the top end of the string

which will improve the string efficiency.

(ii) Capacitance Distribution Network

Self capacitance of Disc C

Link pin to ground capacitanceC

C = 0.2C 2.0KC

C '

C

C

I1

I3

C

C

V1

I2C

I1

I2

C

I3 I

4 C

V1

V2

V3

V4

ixCx

iy

Cy

(1)

(2)

(3)

: 9 : ESE MAINS


Static Capacitance Cx, Cy

(a) string =

unitany)or(

Discanyacrossvoltageimummax4

VVVV 4321

ix = I2 , iy = I3

with guard wires.

I4 = I3 = I2 V4 = V3 = V2 = V (say)

KCL at metal link (1):

I2 = I1 + I1

V2C = V1C + V1C

V2 = V1 + V1 C

C '

V2 = V1 + V1 0.2

V2 = 1.2V1

V1 =2.1

V

2.1

V2

V1 = 0.833V

V2 > V1:

string =2

4321

V4

VVVV

string = 95.83%4

3.833

V4

VVV0.833V

(b)Total voltage across the string

= 40kV

V1 + V2+V3 + V4 = 40kV

3.833V = 40kV

V = 10.43kV

V2 = V3 = V4 = 10.43kV

V1 = 0.833 10.43= 8.69kV

02. (c)

Sol: Given data,

Nf = 750r.p.m, Vt = 250V, Is = 60A,

Ra = 0.4 , Rsh = 125 and Vb = 2V

Now Is = 6At

shsh

V 250I 2A

R 125

Ia = Is – Ish = 4A

The back e.m.f = Ebo = V–IaRa –Vb

= 250 – 40.4 – 2

= 246.4V

Full load back e.m.f = V– IaRa–Vb

= 250 – [60 –Ish ] 0.4 –2

= 224.8V

b0 0

bf f

E N

E N

No-load speed 0

246.4 750N

224.8

= 822 r.p.m

(ii) for full load speed 600 r.p.m

=bf

bf

E (new) 600

E 750

= bf

224.8 600E (new) 179.84V

750

179.84 = Vt –58(Ra +Rse) – Vb

= 250 – 58 (0.4+Rse) –2

Rse = 0.7752

Ish

Rsh

Is

M

Ia

Ra

Vt = 250V

: 10 : ESE MAINS


(iii) b

bf f f

E N

E N

Eb = Vt – 30 (Ra) – Vb

= 250 – 30 0.4 – 2

= 236V

9008.224

750236

f

= 0.8748

Percentage reduction in flux per pole

=f

1 100

= 12.52%

03. (a) (i)

Sol:

When T1 is ON, Vmsin t = Ldt

di0

tdtsinL

Vdi m

0

ktcosL

Vi m

0

k can be determined by applying initial

condition

At t = i0 = 0

k =

cosL

Vm

cos

L

Vtcos

L

Vi mm

0

tcoscosL

Vi m

0

03. (a) (ii)

Sol: Given data:

Vc (0) = - 100 V

When the thyristor is turned on at t = 0, the

voltage equation for the circuit is

sVidtC1

dtdi

L

Its Laplace transform is

s

V

s

0VC

s

sI

C

1sIsL sc

LC

1s

1.

L

VVsI

2

cos

LC

1s

LC

1LC/1

L

300

2Vmsint+

–

o

+

–

io

i

t

Fig.(b)

t

t

t

i0

/2

L300

0

200V

VC

VL,VC

i0

300V

–100V

/2

VL

: 11 : ESE MAINS


: 12 : ESE MAINS


Its Laplace inverse is

i (t) =LC

1wheretsin

L

300oo

o

tcos300

dt

tdiLV oL

tcos300200VVV oLsC

The current and voltage waveforms are as

shown in Fig. (b). At /2,

VL tends to reverse and as a result, diode D

gets forward and current iL starts flowing

through D as iD . VL is therefore zero from

/2 to . Voltage VC remains 200 V and

current i is zero from /2 to shown in

Fig.(b).

03. (b) (i)

Sol: Advantages of Digital Techniques:

1-Improved performance:

Relay performances can be seen from

different perspectives. Some of these

important aspects are;

More complex and hence improved

signal processing.

Higher selectivity and accuracy leading.

Improved security and availability

through self-monitoring.

Shorter response time.

Faster and more accurate localization of

faults.

Greater flexibility due to the fact that

functionality is determined by software

rather than hardware.

Less different parts simplified

maintenance and spare parts

managements.

2-Long-Term Stability:

Long-term stability of the equipment

influences maintenance and routine testing

and therefore strongly reduces the total

lifetime cost for that equipment. Digital

protection and control systems offer the

following advantages;

No, or only very few adjustment points

within the equipment.

Performance largely unaffected by

component aging.

Possibility of automatic calibration and

adjustment.

Abnormality can be detected by self-

monitoring, reducing risk for

malfunction.

Higher availability.

3-Security and Availability: Protection

relays based digital techniques can fully

utilize self-monitoring and self-diagnosis

elements with approximately 5-10% of the

processing power. As a gain, both the

availability and security are increased at the

same time.

: 13 : ESE MAINS


The availability increases as all major

functional units are tested periodically and

in short intervals.

4-Supplementary and/or associated

functions: One of the important aspects of

the supplementary and/or associated

functions is communication. It is to simplify

monitoring the entire system and to utilize

stored information and capabilities.

In case of disturbance, fault location and

trouble shooting can be simplified much

easier and in a shorter time if information

about the operation of the individual

protection and control devices is available in

time.

5-Cost: All other things being equal, the

cost of a relay is the main consideration in

its acceptability. Over the years, the cost of

digital computers has steadily declined, at

the same time their computational power has

increased substantially. The cost of the

analog relays has steadily increased over the

same period mainly because of some design

improvements. It is estimated that the cost of

the most sophisticate digital relays with all

its abilities and features including software

costs, would be the same as that of analog

relay or even cheaper.

03. (b) (ii)

Sol: Relay current,

A10400

000,4

ratioCT

I,currentFaultI f

g

Pick-up value of relay = Current setting

rated current of secondary of CT

= 1.25 1 = 1.25

PSM of relay =relayofvalueupPick

Ig

825.1

10

Form the given data, the operating time for

PSM of 8 is 4. 2 seconds

Actual operating time of relay = 4.2 TSM

= 4.2 0.6 = 2.52 seconds.

03. (c)

Sol: Given data:

V = 2000V, IL = 100A = IFL, If = 2.5A

Synchronous impedance100

500

I

EZ

5.2Isc

fs

f

Zs = 5

22ss RZX

22 8.05 = 4.935

Case (i)

unity power factor, (U.P.F)

cos = 1, sin = 0

2sa2

aa XIsinVRIcosVE

: 14 : ESE MAINS


22 935.4100020008.010012000

E = 2137.742V

Regulation = 100V

VE

%88.61002000

2000742.2137

Case (ii)

0.8 leading, cos = 0.8 sin = 0.6

2sa2

aa XIsinVRIcosVE

22 935.41006.020008.01008.02000

= 1822.5V

100V

VEgulationRe

gReve%87.8

Case (iii)

0.707 lag, cos = 0.707 and sin = 0.707

2sa2

aa XIsinVRIcosVE

22 935.4100707.020008.0100707.02000

= 2422.9305V

100V

VEgulationRe

1002000

200093.2422

%146.21

04. (a)

Sol: Given that

Vdc = 12 V, V0 = 24 V , I0 = 0.5 A

L = 150 µH, C = 470 µF, fs = 20 kHz

D1

VV dc

0

0

dc

V

VD1

24

121D = 0.5

(i) Peak to peak ripple in inductor current

dcL

VΔI DTL

6 3

12 10.5 2A

150 10 20 10

Average Inductor current IL =

dc2

V

R 1 D

IL0I

1 D

= A10.51

0.5

IL max = IL +2

21

2

ΔIL = 2 A

IL min = IL –2

21

2

ΔIL = 0 A

As ,2

LL

II the given condition is at

boundary

iD

iD peak

t

QiD average=I0

DT

T

t1

(1D)T

: 15 : ESE MAINS


2,, DTL

Vii dc

peakLpeakD A

During off time dc oD V Vdi

dt L

Ddi

dt 6

12 2480,000

150 10

A/s

D,peak oD

1

i Idi80,000

dt t

D,peak o 1

o

i I tQ 1V

C 2 C

66

2 0.5118.75 10 29.92

2 470 10

= 29.92 mV

Note: Expression forC

DTIV o

o is valid

only when the minimum value of iL is

greater than Io in the continuous conduction

mode of operation.

(ii) From (i), it is concluded that the

converter is operating at boundary condition

between continuous and discontinuous.

Therefore, iD,peak is as shown below:

816.0

1

3

2 2

12

,

T

TDI rmsD A

5.0, oavgD II A

RMS value of ripple current of diode

current,

2 2ripple,rms D,rms oI I I

2 20.816 0.5 0.645 A

04. (b)

Sol: Supply voltage, V = 11 kV

Frequency, f = 50 Hz

Inductive reactance, XL = 10

Capacitance between phase to neutral,

C = 0.03 F

10fL2XL

H0318.0502

10

f2

10L

kV98.83

112

3

V2Vm

(i) Maximum restriking voltage = 2Vm

= 2 8.98

= 17.96 kV

(ii) Frequency of oscillation, fn =t2

1

LC2

1

61003.00318.02

1

= 5.15 kHz

(iii) Average RRRV =LC

96.17

t

V2 m

61003.00318.0

96.17

= 185089 kV/sec = 0.185089 kV/-sec.

iD

iD peak

t

iD average=I0

(1D)T

2A

: 16 : ESE MAINS


(iv) Maximum RRRV =LC

Vm

6

3

1003.00318.0

1098.8

= 290 106 V/sec

= 290 V/sec

(v) Time for maximum rate of rise of

restriking voltage is

i.e., LC2

t

= 61003.00318.02

= 48.50 -sec

04. (c)

Sol: Given data,

Ra = 1, Rse = 0.5, Rsh = 100, P0 = 4kW,

Vt = 200V and Vb = 2V

(Consider winding is wave winding)

P0 = Vt IL

IL = 4000/200=20A

(i) For short shunt:

Ia = Ish +IL

The voltage across shunt field is

= 200 + ILRse

= 200 + 20 0.5

= 210V

Ish =100

210

R

210

sh

= 2.10 A

Ia = 2.10+20=22.1 AThe generated emf

Eg = IaRa + ILRse +Vt +Brush drop

= 22.10 + 10 + 200 + 2

= 234.10 V

(ii) For long shunt:

Ia = IL + Ish

The voltage across shunt field is 200V

sh

200I 2 A

100

Ia = IL+Ish = 20+2 =22 A

The generated e.m.f

Eg = IaRa + IaRse+ Vt + 2

= 22 + 22 (0.5) + 200 + 2

= 235 V

(ii) Given data, Z = 200, A = 4, N = 750rpm

For short shunt generator Eg = 234.10 V

g

ZNPE

60A

P

234.12 60 493.64 mWb

200 750 4

For long shunt Generator Eg = 235V

P

235 60 494

200 750 4

mWb

Ish

Rsh

IL

G

Ia

Ra

Eg = 240VVt

Rse

LOAD

Ish

Rsh

IL

G

Ia

Ra

Vt= 200V

Rse

LOAD

: 17 : ESE MAINS


: 18 : ESE MAINS


05. (a) (i)

Sol: Advantages of HVDC transmission

(i) The line construction is simpler,

cheaper.

(ii) Power per conductor is more and also

power per circuit is same

(iii) The charging current is totally absent.

So that the length of HVDC

transmission line is not limited.

(iv) No skin effect in HVDC line. As a

merit of this, conductor cross-section is

fully utilized.

(v) HVDC line operates at unity power

factor and charging currents are absent

therefore no reactive power

compensation is required.

(vi) Less corona loss and radio interference

because frequency is zero in case of

D.C.

(vii) The level of switching surges due to

DC is lower as compared to AC and

hence, the same size of conductors

and string insulators can be used for

higher operating voltages.

(viii)No stability problem in HVDC.

Operational problems of HVDC

Transmission:

(i) The converters required at both ends of

the HVDC line more expensive and

these converters have very little

overload capacity and they absorb

reactive power which must be supplied

locally

(ii) The converters produce lot of

harmonics DC and AC sides which

may cause interference with audio-

frequency communication lines.

(iii) Voltage transformation is not easier in

case of DC.

(iv) Circuit breaking for multi-terminal is

difficult.

05. (a) (ii)

Sol: The voltage boost due to a shunt capacitor is

distributed over the transmission line where

as the change in voltages between the two

ends of the series capacitor where it is

connected, is sudden.

Let 1cQ be the reactive power of the shunt

capacitor, V be the receiving end voltage

and X be the reactance of the line. The

current through the capacitor will be

V

Q1c

and the drop due to this current in hv line

will be

V

Q1c X.

Similarly let Qc be the reactive power of the

series capacitor, I be the line current and

sin be the sine of the power factor angle of

the load. The voltage drop across the series

: 19 : ESE MAINS


capacitor will be

I

Qc sin , since the

magnitude of the voltage across the

capacitor is

I

Qc .

For a typical load with p.f 0.8 lag,

sin r = 0.6 and assumeV

IX = 0.1

For equality of voltage boost with both

shunt and series capacitors.

V

Q1c X =

I

Qc sin r

1.0

6.0

V

IXsin

Q

Q

c

1c

= 6.

It is evident that for same voltage boost,

reactive power capacity of a shunt capacitor

is greater than that of a series capacitor.

Therefore, the shunt capacitor improves the

p.f. of the load where as the series capacitor

has little effect on the p.f.

05. (a) (iii)

Sol: Let the over head transmission line is

connected through the cable of surge

impedances are Z1,Z2 respectively. When a

wave travels over the line and enters the

cable, then it suffers reflection and

refraction at the junction ‘x’.

Let V and I are the Incident voltage and

current waves. 11 IandV are the reflected

voltage and current waves, 1111 IandV are

the transmitted voltage and current waves

respectively.

Refracted or transmitted wave

= Incident wave + Reflected wave

But we know that

2

1111

1

11

1 Z

VIand

Z

VI,

Z

VI

111 III …………. (1)111 VVV …………. (2)

From equation (1),1

1

12

11

Z

V

Z

V

Z

V

VVVZ

VV

Z

V

Z

V 111

1

11

12

11

1112

11

Z

1

Z

1V

Z

1

Z

1V

1 2

1 1 2

Z Z2V

Z Z Z

21

211

ZZ

VZ2V,voltagefractedRe

..... (3)

From equation (1),1

1

12

11

Z

V

Z

V

Z

V

111

1

1

12

1

VVVZ

V

Z

V

Z

VV

2121

1

Z

1

Z

1V

Z

1

Z

1V

: 20 : ESE MAINS


2 1 2 11

1 2 1 2

Z Z Z ZV V

Z Z Z Z

Reflected Voltage, 12

121

ZZ

ZZVV

…(4)

Given that Incident voltage, V = 100 kV

Surge impedance of the over head line,

400Z1

Surge impedance of the cable, 40Z2

From equation (3), Refracted or transmitted

voltage

21

211

ZZ

ZV2V

kV18.18

40400

401002

Reflected voltage, 12

121

ZZ

ZZVV

40040

40040100

kV81.81

05. (a) (iv)

Sol: Air density factor

9767.020273

73392

t273

b92.3

Phase-to-neutral critical disruptive voltage

will be

r

dlnmrgV 0c

=1

400ln9767.096.01.210.1

kV53.118

And

Line phase voltage kV02.1273

220Vp

Since VP is greater than Vc corona will be

present. Using Peek’s formula for corona

loss in a fair weather, we get

2cP5

c VVd

r25f10241P

= 67.0 kW/phase/km

For rainy weather, the value of critical

disruptive voltage will be taken as 0.8Vc.

Thus the loss due to corona will be

2cp5

c V8.0Vd

r25f10241P

km/phase/kW59.9

05. (b)

Sol: Per phase voltage, .V38103

6600Vt

Per phase armature current,

.A26266003

103000I

3

a

Percentage reactance,

.100I/V

ohmsinXX

at

ss

Xs in ohms = .90.2262

3810

100

20

(i) At no load, ≅ 0 and Vt = Ef.

Synchronizing power per mechanical

degree,

360

P.cos

X

EV3

360

P.

d

dP.mP

s

fts

: 21 : ESE MAINS


.kW1048

360

8

90.2

38103 2

The corresponding synchronizing torque is

s

ss

ss n2

PP.

N2

60T

31 81048 10

2 100

13,340 Nm.

(ii) satf ZIVE

fE 3810 j262 0.8 j0.6 2.9

o4310 8.11

The synchronizing power per mechanical

degree is

cosX

V.E.

360

P3P

s

tfs

kW117311.8cos

29

38104310.

360

83

The corresponding synchronizing torque is

s ss

1T P

2 n

381173 10 14,930 Nm.

2 100

05. (c)

Sol: (i) Inductive reactance = njLjn 87.11 Ω

Capacitive reactance =n

j

Cjn

68.231

Ω

22 1

Z R n Ln C

22 23.68

10 11.87nn

10

68.2387.11

tan 1 nn

n

dco

n 1,3,5...

4Vv sin n t

n

280.1sin 377t 93.4sin 3 377t

56.02sin 5 377t

40.01sin 7 377t 31.12sin 9 377t

47.151 Z Ω And 74.491

465.293 Z Ω And 16.703

52.555 Z Ω And 63.795

33.807 Z Ω And 85.827

67.1049 Z Ω And 52.849

oi 18.1sin 377t 49.7

3.17sin 3 377t 70.17

1sin 5 377t 79.63

0.5sin 7 377t 82.85

0.3sin 9 377t 84.52 A

(ii) Peak value of the fundamental load current

= 18.1 A

RMS value of the fundamental load current

= 798.122

1.18 A

(iii) RMS value of load current

=2 2 2 2 2

18.1 31.7 1 0.5 0.3

2 2 2 2 2

13.02 A

: 22 : ESE MAINS


% THD in load current =2 2or o1

o1

I I100

I

2 213.02 12.798100 19%

12.798

(iv) Power delivered to the load at fundamental

frequency,

163810798.12 2211 RIP oo W

Net power delivered to the load,

2.16951002.13 22 RIP oro W

(v) Average dc input current = 7.7dc

o

V

P A

(vi) Peak current in switch

= 41.18202.13 A

RMS value of the switch current

206.92

41.18 A

: 23 : ESE MAINS


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