: 2 : ESE MAINS - ACE Engineering Academy - Leading ... 2 : ESE MAINS ACE Engineering Academy...
Transcript of : 2 : ESE MAINS - ACE Engineering Academy - Leading ... 2 : ESE MAINS ACE Engineering Academy...
: 2 : ESE MAINS
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
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
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
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
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
: 5 : ESE MAINS
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
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
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
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
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
(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
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
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
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
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
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
(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
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
: 12 : ESE MAINS
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
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
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
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
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
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
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
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
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
(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
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
: 18 : ESE MAINS
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
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
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
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
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
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
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
.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
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata
% 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