ee volume - 01 corrections€¦ · torque angle : it is the angle between rotor field axis and...

GATE2019(PreviousYear)EEVolume–01

Chapter1:ElectricalMachine1.6 Page Number 1.2

A 50 Hz, 220 V/440 V, 5 kVA single-phase transformer operates on 220 V, 40 Hz supply with secondary winding. Then

1.19 Page Number 1.21 45.64kVA,500W,500W Maximum efficiency occurs when, Variable loss = Constant loss

2

flcu im P P

Where, 2

flcum cuP m P = Copper loss at fraction of load (m)

and flcuP Full load copper loss (m = 1)

500

600fl

i

cu

Pm

P

0.9128m Loading at maximum efficiency condition is given by,

max

S

0.9128 50 kVAm S

max

S

45.64 kVA

Since, iron losses are independent of loading. Therefore, iron and copper losses at this point are,

500 Wi cumP P

Hence, the load at which maximum efficiency occurs is 45.64 kVA and the iron and copper losses are 500 W and 500 W respectively.

Key Point

flcu iP P at full load 1m

At maximum efficiency,

2

flcu im P P at any load m

1.55 Page Number 1.39 . Method 2 :

No load current Linear dimension

1 1

2 2

dimension

dimension

I

I

2

0.5 1

2I

2 2 0.5 0.7 AI

Power (Linear dimension) 3

3

1 1

2 2

dimension

dimension

P

P

3

2

55 1

2P

2 2 2 55P

2 1.414 110 155.6 WP

Hence, the correct option is (B). 1.67 Page Number 1.49 For parallel operation of 1T and 2T load shared by transformer 2T is given by,

12

1 2L

ZS S

Z Z

02

1 6(1000 36.87 )

1 6 0.8 4.8

jS

j j

02

1 6(1000 36.87 )

1.8 10.8

jS

j

02

1 6(1000 36.87 )

1.8(1 6)

jS

j

02

100036.87

1.8S

02 555.55 36.87S

2.9 Page Number 1.64 (iii) Hence,

2 1CA A C CB CBV V V V

2.16 Page Number 1.66 Given : (i) 110/220 V, Y - 3-phase transformer

3.52 Page Number 1.79 A 250 V DC shunt machine has armature circuit resistance of 0.6 and field circuit resistance of

125 . The machine is connected to 250 V supply mains. The machine is operated as a generator and

then as a motor separately. The line current of the machine in both the cases is 50 A. The ratio of the speed as a generator to the speed as a motor is ______. [Set - 02]

3.29 Page Number 1.96 Torque, ' se ak IT

3.39 Page Number 1.102

(A) Armature EMF, 60

P NZE K

A

where, 2

PZK

A

= constant

2

60

N


Solution changed

Total copper losses 2 2sh sh a aI R I R

2 25 44 50 0.01 1125 W

Also, stray losses 375 W (given)

Hence, total loss 1125 375 1500 W

and output power,

220 45 9900 Wout LP VI

% Efficiency, 100

out

out losses

P

P P

9900

% 1009900 1500

86.84%

Hence, the percentage efficiency of the DC generator is 86.84 %.


At rated voltage for DC series motor, (figure (a))

1 11 a a a seV I R I RE

1 220 (30 0.4) (30 0.1) 205 VE

Hence, 2

1

0.5 15

30

E N

E N

2

1

1

4

E

E

12

20551.25V

44

EE

At reduced load, (figure (b))

22 a totalE V I R

total ext a seR R R R …(iii)

2

2total

a

V ER

I

220 51.25

11.2515totalR


Where Torque angle

Torque angle : It is the angle between rotor field axis and resultant flux density wave.

Hence, the correct option is (B).

4.12 Page Number 1.137 For alternator,

( ) ( )in out Cu loss mech lossP P P P

23 cos 3 0in t a a aP V I I R

3 215 10 3 415 0.9 3 0.5a aI I

21.5 646.9 15000 0a aI I

2646.9 (646.9) 4 1.5 ( 15000)

2 1.5aI

22.059AaI or 453.327A

Ignoring negative value of aI

22.059AaI

Now, excitation voltage for a synchronous generator at a leading pf is given by,

2 2( cos ) ( sin )f t a a t a sE V I R V I X

2 2 2( cos ) ( sin )f t a a t a sE V I R V I X

2

2400(239.6 0.9 22.059 0.5)

3

2(239.6 0.435 22.059 )sX

2 2 2230.94 (226.66) (104.226 22.059 )sX

21958 (104.226 22.059 )sX

(44.255) 104.226 22.059 sX

104.229 44.255

22.059104.229 44.255sX

22.059 148.484, 59.974sX

6.731 , 2.71sX

Generally the reactance should be low so practically 2.71 is preferable. Hence, the synchronous reactance is 2.71 .

4.24 Page Number 1.122 A single-phase, 2000 V alternator has armature resistance and reactance of 0.8 ohms and 4.94 ohms

respectively. The voltage regulation of the alternator at 100 A load at 0.8 leading power factor is

Stator mmf

Rotor mmf

Resultant mmfand flux density


Also,

About phase sequences :

(i) The prime mover of the incoming machine is started and brought near to rated speed. The field current of incoming machine is adjusted so that it become equal to infinite bus voltage. The three lamps flicker at a rate equal to the difference in frequencies of incoming machine of infinite bus.

(ii) If the phase sequence are properly connected then all the lamps will be bright and dark at the same time. If this is not the case then it means that the phase sequence are not correct. In order to correct the phase sequence, two leads of the line of the incoming machine should be interchanged.

About frequency :

Case I : If incoming generator is fast.

'f f then relative speed will be ( ' ) i.e. we can assume , ,R Y B are stationary where ', ', 'R Y B are

moving with ( ' ) in the anticlockwise direction then - 'R R will coincide first aL become dark, after

that - 'Y Y will coincide so bL becomes dark and finally - 'B B will coincide so cL becomes dark.

So sequence will be - -a b cL L L

cL

bL

aL

R

'R

B 'B Y

'Y

'f

f

Case II : If incoming generator is slow.

We can assume , ,R Y B are stationary whereas ', ', 'R Y B are moving relatively ( - ') in clock wise

direction. ' -B B will coincide first so cL will become dark. ' -Y Y will coincide second so BL will become

dark and at last ' -R R will coincide so aL will become dark.

So sequence will be - -c b aL L L

Hence, the correct option is (A).


4.77 Page Number 1.169 Let ‘f ’ is a common frequency of operation.

Applying property of similar triangles,

5.3 Page Number 1.187 Case I : Delta connection

R

'R

B 'B Y

'Y

cL

bL

aL

5.20 Page Number 1.175 & 1.194 A 3-phase, 20 kW, 400 V, 1470 rpm, 50 Hz squirrel cage induction motor develops a torque of 100 N-m

at a speed of 1400 rpm. If the motor is connected to a 30 Hz supply, for keeping the same air-gap flux, the supply voltage should be _____V and for the same load torque, the new speed will be ________ rpm.

(A) 240 V and 864 rpm (B) 400 V and 800 rpm

(C) 240 V and 1400 rpm (D) 400 V and 1400 rpm

Solution :

Now, for constant air gap flux.

Assume same flux i.e. /V f k

1 2

1 2

V V

f f

2400

50 30

V

2

400 30240 V

50V

Slip, 1

1500 14000.0667

1500s

For same load torque ( 1 2T T ),

22

22 22 2

3

( )es

sET R

R sX

21

1eT sV

f

Since, 2 2

1 1 2 2

1 2

s V s V

f f

1 2

1 2

s s

f f

20.0667

50 30

s

2 0.04s

Slip, 22

2

0.04 s m

s

N Ns

N

New speed can be calculated as,

22

120 120 30

4s

fN

P

2 900 rpmsN

2 (1 0.04) 900(1 0.04)m sN N

864 rpmmN

Hence, the correct option is (A).


The equivalent circuit of the induction motor referred to stator is shown below,

r rr r

R RR R

s s

Rotorresistance Load Resistance

1rr r

R sR R

s s


In previous problem (Q. 5.51) already calculated ( ) 2.45st DOL flT T .



(D) to increase the stored energy at constant flux.


The self-inductance of rotor, 2r

rr

NL

Rl

Self-inductance of stator, 2s

ss

NL

Rl

Chapter2:PowerElectronics1.22 Page Number 2.13

Given :


Given configuration is shown below,

When switch is ON.

1 OFFD

2 OND

3 OFFD

9 1 1

gIR

3(max)

7100 10gI

R

3

770

100 10R

[Not in option]

3(min)

7150 10gI

R

3

746.7

150 10R

Hence, the correct option is (C).

RPT

10 V

CEV

K

1.0

200 V

SCR L

1V 2V

2D

3D C

G

Switch

1D

RPT

10 V

ON (1 V)

9 V 9 V

1V

1 VgkV

gI


For turning ON the SCR,

L AI I

Applying KVL, thyristor current is,

199

(1 )Rt

LA

L

I eR

3

13 150 10

199250 10 (1 )

1

t

e

1000

3 1501.25628 10 1t

e

6.667 31 1.25628 10te

6.667 0.9987te

6.667 1

0.9987te

6.667 ln1.00125t

188.55 secpt

Maximum pulse width [ V- sec ] of transformer

10 188.55 1885.5 V - sec

Which is very close to option (A).


Device (B) :

When thyristor is in forward conduction mode,

0sV , 0si


A single phase fully controlled converter bridge is used for electrical braking of a separately excited dc motor. The dc motor load is represented by an equivalent circuit as shown in the figure.


Output voltage and current waveform of this single phase diode bridge rectifier is shown below,

200 VSV

1LR AI

150 mHL

2.10 Page Number 2.33 (iii) Back emf constant, bk 0.5V/rpm,

2.18 Page Number 2.39 Charging current flows during 1 1t

1 1

110 ( )1

( cos )2 m E tI V t

R

2.30 Page Number 2.45 Input power factor is given by,

IPF 00.9 cos3 00 .779 Hence, the input power factor of the bridge is 0.779. 2.32 Page Number 2.46 During positive half cycle ( )t ,

Thyristor 1T and diode 2D are in forward bias condition, at 030 , thyristor 1T is fired so output voltage

0 sV V .

During negative half cycle ( 2 )t ,

Diode 3D and 4D are forward biased so output voltage 0 sV V .


Given : A Buck converter

1 MHz, 0.5, ?f D I

. Method 1 :

For type A chopper

0 6 0.5 100 50 VSV V

When switch is ON, apply KVL

0 0S

dVV L V

dt

0S

dVV V L

dt

0SON

V VT I

L

3 3

(100 50) 0.50.125 A

200 10 1 10I


. Method 2 :

Ripple current for step down / type A chopper is given by,

/ /

/

(1 ) (1 )

1

ON OFFT TS

T

V e eI

R e

Where, 3

3200 1040 10

5

L

R

311 10

1000T

30.5 10ONT DT

3(1 ) 0.5 10OFFT D T

Therefore,

3 3

3 3

3

3

0.5 10 0.5 10

40 10 40 10

1 10

40 10

1 1

1

e e

I

e

0.125 AI


200 mH

5 �i

100 V

. Method 3 :

For step down chopper, maximum value of inductor / load current is given by,

3

3

3

3

0.5 10/ 40 10

max / 1 10

40 10

1 100 1

1 51

ONTS

T

V e eI

R ee

max 10.062 AI

Minimum value of inductor / load current is given by,

3

3

3

3

0.5 10/ 40 10

min / 1 10

40 10

1 100 1

1 51

ONTS

T

V e eI

R ee

min 9.937 AI

Therefore, ripple current is given by,

max min 10.062 9.937I I I

0.125 AI


. Method 4 :

For type A chopper

When D = 0.5 maximum value of ripple current occurs.

max 3 3

100

4 4 1 10 200 10SV

IfL

max 0.125 AI



(i) Turn-off time of each SCR = 50 sec.

(ii) Safety margin (SM) = 2


Now, Input power s sV I

4000 360 sI

(avg) 11.1 As sI I

As frequency is not given so ripple cannot be calculated so we can assume that the case is continuous conduction ripple free.


Average value of inductor current is,

0 1537.5

1 0.4L

II

D

…(ii)


A three-phase voltage source inverter supplies a purely inductive three-phase load. Upon Fourier analysis

the output voltage waveform is found to have nth order harmonics of magnitude na times that of the

fundamental frequency component ( 1)na , the load current would then have an thn order harmonic of

magnitude


The range of firing angle for controlled operation,

0180

The maximum firing angle at which the voltage across the device becomes zero or for continuous

conduction of thyristor 0max 45 because if we provide firing angle less than 045 , switch will not be

fully controlled. If we provide 045 then voltage across switch will not be zero.

. Method 1 :

RMS value of current through SCR

21

sin( )2rms

mT

VI t d t

Z

2

2

1 cos 2( )( )

2 2rms

mT

V tI d t

Z

( )2

2

sin 2( )

2 2 2rms

mT

V tI

Zt

Chapter3:PowerSystemAnalysis3.3 Page Number 3.36

The power system can be represented as shown below,

From active power balance, R LP P

From reactive power balance,

R C LQ Q Q

C RQ Q

Therefore, reactive power injected at load is

( 10.1) 10.1 MVARC RQQ

3.4 page Number 3.37

. Method 2 :

Apparent power at receiving is given by,

R R RS P jQ


Consider the two bus power system network with given loads as shown in the figure. All the values shown in the figure are in per unit. The reactive power supplied by generator 1G and 2G are

1GQ and 2GQ

respectively. The per unit values of 1GQ ,

2GQ and line reactive power loss loss( )Q respectively are


. Method 1 :

At Generator 1 ( 1G ) :

Active power balance at input side is,

20 15 5SP sinS R

L

V V

X

01 15 sin 30

0.2

Sending end reactive power is given by,

2

cosS S RS

V V VQ

X X

2

01 1 1cos30 1.34 pu

0.2 0.2SQ

So, reactive power balance at input side is,

1 5 1.34 6.34 puGQ

120 GjQ

S SP jQ

1 1P jQ15 5j

1

1.34SQ 1GQ

1 5Q

At Generator 2 ( 2G ) :

Receiving end reactive power is given by,

2

cosR S RR

V V VQ

X X

2

01 1 1cos30 1.34

0.1 0.1RQ

So, from reactive power balance at output side,

2 2G RQ Q Q

2 2 10 ( 1.34)G RQ Q Q

2 11.34 puGQ


Note : Formula of RQ is designed for towards receiving end but we are getting negative values so to

make it positive we revered the arrow to calculate 2GQ .

. Method 2 :

Active power flow in the line is given by,

sinR SS R

V VP P

X

01 15 sin 30

0.1

So, current flowing through transmission line will be,

0 0

0

1 30 1 0

0.1 90S R

SS

V VI

X

05.1763 15 puSI

Sending end apparent power *S SV I

0 0 01 30 5.1763 15 5.1763 15

RQ

215 GjQ

2 2P jQ20 10j

2GQRQ

2Q

SI

RI

5 puSP

1 01 0

0.1j

5 pu, 1.34 puS SP Q

Receiving end apparent power *R RV I

0 0 01 0 5.1763 15 5.1763 15

5 pu, 1.34 puR RP Q

(i)

1 5 1.334 6.334GQ

(ii)

2 10 1.34 11.34GQ

(iii)

Loss absorbedQ Q

1.34 1.34 2.68 puS RQ Q



. Method 2 :

1 kW 2 kWLP P

1 kVAR 2 kVARLQ Q

1 kVARCQ

Receiving end p.f. after compensation is,

2 2 2 222 1

cos L

L L C

P P

P Q P Q Q

SQ1GQ

1Q

2GQRQ

2Q

1.34SQ ' 1.34RQ

1 2Q

CQ

2

1Q

So, LP should be low, LQ should be high.

2 22

1 1cos 0.707 lag

21 2 1

OR

If we can’t imagine then we can put different given values to get the exact answer.

For 1 kW, 1 kWL LP Q

2 2 2

1cos 1

1 (1 1)


2 2 2

1 1cos

21 (2 1)


2 2 2

2cos 1

2 (1 1)


2 2 2

2 2cos 0.854

52 (2 1)



Transmitted voltage is given by,

T v sV B V

where, 2

1 2

2v

Z

ZB

Z


OR

TV towards C,

Transmitted voltage is given by,

2

1 2

2 2 80250

480T I

ZV V

Z Z

83.33 kVTV

AB C

2 80Z 1 400Z

250 kVIV

TransmissionLine

Cable


OR

RV towards A,

Reflected voltage is given by,

2 1

1 2

80 400250

480R I

Z ZV V

Z Z

166.67 kVRV


OR Current in cable BC, Transmitted current is given by,

3

2

83.33 101.042 kA

80T

T

VI

Z

5.8 Page Number 3.55 Given figure is shown below,

For pure inductor, transmitted voltage is given by,

/2 Ct ZT IV V e

When the waves touch, the discontinuity occurs i.e. 0t

02T IV V e

2 2 puT IV V



Each conductor of a 33 kV, 3-phase system is suspended by a string of three similar insulators. The ratio

of shunt capacitance to mutual capacitance is 0.1.

7.10 Page Number 3.63 The voltage (in kV) at the line end of the unit will be ______.

A BIV

166.67 kVRV 83.33 kVTV

C

L

1.0 pu

Common Data for Questions 7.5 & 7.6


Condition Load impedance Loading

Leading : 2 21 1

2 2LCV LI 0LZ Z L SILP P

Lagging : 2 21

22

1LCV LI 0LZ Z L SILP P

7.9 Page Number 3.67 Maximum electric stress is given by,

2

36.6 10

31.5

ln 0.75 ln0 7

1.

05

Vg

Rr

r


1line : ,22 10 k 5 h sV 0 o mlX

8.6 Page Number 3.76 The new base value of different components is as shown below,

9.1 Page Number 3.80 Fault impedance,

1 1 2( )f g gZ X X X

1 0.1 (0.04 )fZ X

1

0.1(0.04 )

0.1 0.04fZX

X

…(i)

From equation (i) and (ii),

0.1(0.04 )

0

1

1 . 0.05 1 4

X

X

0.004 0.1

0.14

X

X


Given :

(i) Power system network :

(ii) 6 MVABASES

Grid

Feeder 1

Feeder 2 CB Load

T

(iii) 3- transformer :

6 MVATS , TZ (0.01 + j0.08) pu

(iv) Feeder : 1 2 (9 18)feeder feederZ Z j

(v) 3- load :

V 11 kV, loadS 5800 kVA,

cos 0.8 lag, Z j0.2 pu

. Case 1:

If we consider load is static then,

To find : MVA rating of circuit breaker.

Impedance of each feeder in per unit for new bases is given by,

2

(New) (Old)(New) (Old)

(Old) (New)

b bpu pu

b b

S VZ Z

S V

1(pu) 2(pu)feeder feederZ Z

1(pu) 2

6(9 18)

(33)feederZ j

1(pu) (0.0495 0.099) pufeederZ j

The equivalent impedance of feeder is given by,

(pu) 1(pu) 2 (pu)||feeder feeder feederZ Z Z

(pu)

0.0495 0.099

2feeder

jZ

(pu) (0.025 0.05) pufeederZ j

For load, 11 kV =1puV

0 00 1 0V V pu (reference)

Load, 5800kVA =5.8MVAS

(pu)

5.80.967 pu

6b

SS

S

Also, (pu) pu puS V I

0.967 1 puI

0.967 puI

Load pf (cos ) 0.8lag

036.87

The current flowing through the load is given by,

00.967 36.87 puI

The impedance diagram of the given system is represented as,

Applying KVL in the above figure ,

( )Leq TE V I Z Z 0 01 0 0.967 36.87E [(0.025 0.05) (0.01 0.08)]j j

01.11 4.15 puE

For solid fault, i.e. fault impedance 0.Z

Post fault equivalent circuit is given by,

Post fault current through CB is given by,

01.11 4.15

0.025 0.05 0.01 0.08fIj j

08.24 70.8 pufI

Fault level to be interrupted by CB is given by,

( . .) pu =1 8.22 puf C B f fS V I

So fault MVA to be interrupted

(C.B.) (C.B.) pu 8.22 6f f bS S S

(C.B.) 49.44 MVAfS

Hence, the MVA rating of the circuit breaker is 49.44 MVA.

. Case 2 :

If we consider that load is motor because it reflects the rating of synchronous motor then,

6 MVABASES , 33 kVBASEV at grid side

E

feederZTZ

(0.01 0.08) puj�

(0.025 0.05) puj� CB 00.967 36.87I � � �

�

�

LOAD

1 pu

01.11 4.15E � �

feederZ

CB

�

�fI

(0.025 0.05) puj�

TZ

(0.01 0.08) puj�

29 18fZ j

19 18fZ j

0.01 0.08TZ j

0.2Z j

mE

gridV

LpuI

(feeder 1 or 2) 2

9 18 (9 18) 6

33puBASE

j jZ

Z

(feeder 1 or 2) 0.0495 0.099 pupuZ j

( ) ( 1) ( 2)||pu feeder pu feeder pu feederZ Z Z

( )pu feederZ0.0495 0.99

0.025 0.05 pu2

jj

For load 11

11 kV, 1 pu11L puV

Load, 5800 kVA 5.8 MVAloadS

( ) 5.80.967 pu

6L load

L puBASE

SS

S

L pu L pu L puS V I

00.9670.967 36.86

1L puI

Pre-fault condition :

[0.01 0.08 0.025 0.05]grid L pu L puV V I j j

0 01 0 0.967 36.86 0.035 0.13j

01.10540 4.164

[ 0.2]m L pu L puE V I j 00.89742 9.928

Now the fault occurs at the load side to find the rating of CB [To get maximum rating, assume fault occurs in between T and CB]

f grid mI I I 0.2

grid m

feed T

V E

Z Z j

0 01.10540 4.164 0.89742 9.928

0.035 0.13 0.2fIj j

0.025 0.05j

0.01 0.08j

0.2j

mE

gridV

LpuI

gridI

mE

CBmI

0.2jZT

feederZ

gridV

f m gridI I I

0 .025 0 .05j

0.01 0 .08j

012.324 80.985fI

1 12.324in pu f pu f puSC V I 12.324

MVA in pu BASESC SC S

MVA 12.324 6SC 73.944 MVA

Hence, the MVA rating of the circuit breaker is 73.944 MVA.

Theory - Page Number 3.89

6. Relationship between line and phase base values in 3- transformers :

(b) - Connection :

( ) ( )b line b phaseV V ( ) ( )3b line b phaseI I

10.3 Page Number 3.90 (A) 0.8 0.6j (B) 0.75 0.22j

(C) 0.75 0.25j (D) 1.5 0.25j

10.5 Page Number 3.96 RYB, BRY and YBR represents the same phase sequence.


Key Point In volts :

1 1

03 30ab anV V or 1

1

0303

aban

VV

2 2

03 30ab anV V or 2

2

0303

aban

VV

In per unit :

1 1

1

0

(rated) (rated)

3 30( )

3

ab anab pu

L p

V VV

V V

1

1

0

(rated)

( ) 30anab pu

p

VV

V

1 1

0( ) ( ) 30ab pu an puV V

or 1 1

0( ) ( ) 30an pu ab puV V

Similarly,

2 2

0( ) ( ) 30ab pu an puV V

or 2 2

0( ) ( ) 30an pu ab puV V

Per unit line voltage = Per unit phase voltage


To find : Z

abc abcf K A f

11pno abcf A f

K

… (i)

where,

2

1 2

11

13

1 1 1

A

0.5 0 0

0 0.5 0

0 0 0.5p no p noV I

… (ii)

From equation (i) and (ii),

1 1

0.5 0 01

0 0.5 0

0 0 0.5abc abcA V A I

K

1

0.5 0 0

0 0.5 0

0 0 0.5abc abcV A A I

Here,

2

2 2

2

1 1 1 0.5 0 0 11

1 0 0.5 0 13

1 0 0 0.5 1 1 1

1 0.5 0.5

0.5 1 0.5

0.5 0.5 1

So, 1 0.5 0.5

0.5 1 0.5

0.5 0.5 1abc abcV I


10.19 page Number 3.103

and zero sequence network for transformer 2T is,

Theory – Page Number 3.106

3. Line to line (LL) fault :

0 0aI , 1 21 2

aa

Fa

EI I

Z Z Z

Fault current, 11 2

33 a

F aF

j EI j I

Z ZZ


Given :

(i) Alternator :

30 MVA,gS 13.8kVbV

1 0.25pu,gX 2 0.35pugX and

0 0.10 pugX

(ii) LLG fault.

To find : Alternator line current.

The sequence network for a LLG fault is given by,

0

11 2 0

1 0

|| 0.25 0.35 || 0.1a

a

EI

X X X j j

0

1

1 03.05

0.25 0.077aI jj j

20 1

2 0

0.353.05

0.35 0.1a a

X jI I j

X X j j

0 2.372aI j

03 3 ( 2.372) 3.116f aI I j j

Magnitude of fault current in pu is 7.116 pu.

Alternator line currents are for LLG fault

�+

aE

1Z

1aV

+

�

2Z2aV

+

�

1aI2aI

F F

0Z 0aV

+

�

0aI

F

3 FZ

bI

0aI

cI

nI

03f b c aI I I I

0aI

1 2 0b b b bI I I I

21 2 0b a a aI I I I

0 01 240 3.05 90bI 0 01 120 0.6777 90 2.372 90

04.804398 132.217bI

1 2 0c c c cI I I I

21 2 0c a a aI I I I

4.8043 47.78277cI

Hence, the alternator line currents is 4.8043 kA.


6

3

100 106.66

3 25 10F LG

I

15396.007 A

Hence, the fault current immediately after the fault is 15396.007 A.


Given : (i) 3- Y- connected generator :

125 MVA, 15.5 kVg tgS V

1 0.1 puX j , 2 0.05 puX j ,

0 0.01 puX j

(ii) Grounding impedance, 0.01 punX j

For fault in phase b, sequence network will be,

0.01 puj

bI

b

a

c

From the sequence network, sequence current is given by,

11 2 0

1 2 0

/

3b b

b b bn

V VI I I

jX jX jX j X

For SLG fault, current is given by, 1 2 03 3 3f b b bI I I I

So, 1 2 0

3

3b

f pun

VI

j X X X X

0

00

3 1 12015.789 210

0.19 90f puI

(kA)f f pu BaseI I I

6

(kA) 3

125 1015.789

3 15.5 10fI

(kA) 73.51 kAfI

Hence, the magnitude of fault current for a b-phase to ground fault is 73.51 kA.


Critical clearing angle,

max 0 max 2 0 max 3 max

max 3 max 2

( ) cos coscos mech

cr

P P P

P P

where, max1 :P Maximum power that can be transferred under pre-fault conditions.

max 2 :P Maximum power that can be transferred under fault conditions.

max 3 :P Maximum power that can be transferred under post-fault conditions

1 0.1jX j

1bI

1bVbV

2 0.05jX j

2bI

0 0.01jX j0bI

3 0.03nj X j


The critical angle , before which breakers 3B and 4B must open so that synchronism is not lost is

Answer table - Page Number 3.133

12.16 C 12.17 1.22 12.18 A 12.19 C 12.20 C


Given :

(i) Lossless alternator

(ii) max 100 MWP

(iii)0

50 MWmechP

(iv) 1

40 MWmechP

Initially

150 MWmP

100sineP

1m eP P

050 100sin

00 30 or 0.523 rad

Now the prime mover abruptly increased by 40 MW so now,

2

50 40 90 MWmP

2m eP P

190 100sin

11 sin (0.9)

01 64.158 or 1.1197 rad

2 1 1.1197

02 115.842 or 2.0202 rad

Accelerating area is given by,

1

2

0

1 ( )m eA P P d

1 1 0 max 0 190 cos cosA P

1 53.703 43.0134 10.68955A

max 100 MWP

2mP

150 mP

0 1 2

Decelerating area is given by,

2

2

1

2 ( )e mA P P d

2 max 1 2 2 1cos cos 90A P

2 100 cos64.158 cos115.842A 90 2.0202 1.1197

2 87.178 81.045 6.133A

2 1A A

Accelerating area is greater than decelerating area, so synchronism of alternator will be lost.

Hence the correct option is (D).


. Method 2 :

From figure,

dE V I X

0 0 01 0 1 36.86 1.2 90E

01.9697 29.167E

01.9697, 29.167E

max

1.9697 1

1.2

E VP SSL

X

max 1.6414 puP

Hence, the steady state power limit is 1.64 pu.


. Method 2 :

For fixed active power cospu puV I 1 1 0.8 0.8 pu

If excitation is reduced beyond criE then the intersection of mP and eP is not possible. So, minimum value

of excitation is criE .

Where max (new ) 0.8P

E 1.2 pudX 01 0V I

eP

1 pumP

E

1E

criE

2E

1 2criE E E E

'

0.8E V

X

1.2 0.8

' 0.96 pu1

E

Hence, the critical excitation corresponding to operation at stability limit is 0.96 pu.


0

1.11 sin

0.3


Case 3 : When compensator is connected at 75% distance : 3 0.75X X

2

33 3sin

XP

V

2

3 30.7si

5n

VP

X …(iii)


Type 3 :

1

1 1

2 2

(

1 2

)

2 ... ...

p q

p q

pp pqBUS new BUS

qp qq

np n

p q p q pp pq qp qq np nq

ll

q

Z Z

Z Z

Z ZZ Z

Z Z

Z Z Z Z Z Z Z Z Z Z

Z

Z Z


Same Question was in chapter 3 question number 3.8 (Page Number 3.33)

Chapter4:Measurement2.35 Page Number 4.13

A (0-50 A) moving coil ammeter has a voltage drop of 0.1 V across its terminals at full scale deflection.

The external shunt resistance (in milliohms) needed to extend its range to (0 - 500 A) is __________.

[Set - 01]

5.10 Page Number 4.63 If the voltage is unbalanced then positive, negative and zero sequence of voltage and current will flow

through the system. Here the system is 3- , 3 wire i.e. either it can be Y isolated neutral or Delta

connection.

(i) Y No chance of zero sequence of current through phase as well as line.

(ii) No chance of zero sequence of current through line.

So wattmeter will read the total power due to only positive and negative sequence component but the overall power remain constant. So wattmeter reading will remain same. Therefore, the power reading is affected by negative sequence voltage but not by zero sequence voltages.


Note : Positive and sequence components exits in the power calculation but the total power remain same and measured by the wattmeter i.e. reading will not get affected.

6.14 Page Number 4.74 (D) P = 1, Q = 5, R = 6, S = 4

Chapter5:Electromagneticfields1.39 Page Number 5.7

ee volume - 01 corrections€¦ · torque angle : it is the angle between rotor field axis and...

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