Samplepractice Exam 15 March Questions and Answers PDF
Samplepractice Exam 15 March Questions and Answers PDF
Samplepractice Exam 15 March Questions and Answers PDF
Ans : B
Q.2 A salient pole synchronous motor is running at no load. Its field current is switched off.
The motor will
(A) come to stop.
(B) continue to run at synchronous speed.
(C) continue to run at a speed slightly more than the synchronous speed.
(D) continue to run at a speed slightly less than the synchronous speed.
Ans: B
Q.3 The d.c. series motor should always be started with load because
Ans: A
Q.4 The frequency of the rotor current in a 3 phase 50 Hz, 4 pole induction motor at full
load speed is about
(A) 50 Hz. (B) 20 Hz.
(C) 2 Hz. (D) Zero.
Ans: C
Ans: A
Ans: A
Ans: A
Q.8 When a synchronous motor is running at synchronous speed, the damper winding
produces
(A) damping torque.
(B) eddy current torque.
(C) torque aiding the developed torque.
(D) no torque.
Ans: D
Q.9 If a transformer primary is energised from a square wave voltage source, its output
voltage will be
(A) A square wave. (B) A sine wave.
(C) A triangular wave. (D) A pulse wave.
Ans: A
(A) I a . (B) I a2 .
1 1
(C) . (D) .
Ia I a2
Ans: B
Q.11 In a 3 – phase induction motor running at slip ‘s’ the mechanical power developed
in terms of air gap power Pg is
Pg
(A) (s − 1)Pg . (B)
(1 − s ) .
(C) (1 − s )Pg . (D) s ⋅ Pg .
Ans: C
(C) is proportional to r2 .
Ans: B
Ans: B
Q.15 The maximum power in cylindrical and salient pole machines is obtained respectively
at load angles of
Ans: D
Q.18 The relative speed between the magnetic fields of stator and rotor under steady state
operation is zero for a
(A) dc machine. (B) 3 phase induction machine.
(C) synchronous machine. (D) single phase induction machine.
Ans: all options are correct
Q.19 The current from the stator of an alternator is taken out to the external load circuit
through
(A) slip rings. (B) commutator segments.
(C) solid connections. (D) carbon brushes.
Ans: C
Q.20 A motor which can conveniently be operated at lagging as well as leading power
factors is the
(A) squirrel cage induction motor. (B) wound rotor induction motor.
(C) synchronous motor. (D) DC shunt motor.
Ans: C
Ans: B
Ans: C
Q.24 Out of the following methods of heating the one which is independent of supply
frequency is
(A) electric arc heating (B) induction heating
(C) electric resistance heating (D) dielectric heating
Ans: C
Q.25 A two-winding single phase transformer has a voltage regulation of 4.5% at full-load
and unity power-factor. At full-load and 0.80 power-factor lagging load the voltage
regulation will be
(A) 4.5%. (B) less than 4.5%.
(C) more than 4.5%. (D) 4.5% or more than 4.5%.
Ans: C
% R = Vr cos Φ + Vx sin Φ
= Vr
p.f = cos Φ =1 ∴ Φ =00
∴ kVA = kW & kVAR =0
Q.26 In a dc shunt motor the terminal voltage is halved while the torque is kept constant.
The resulting approximate variation in speed ' ω' and armature current ' I a ' will be
Ans: B
N V – IaR or N Eb
T Ia Φ , Φ Ia
∴ T Ia2
Q.27 A balanced three-phase, 50 Hz voltage is applied to a 3 phase, 4 pole, induction
motor. When the motor is delivering rated output, the slip is found to be 0.05. The
speed of the rotor m.m.f. relative to the rotor structure is
(A) 1500 r.p.m. (B) 1425 r.p.m.
(C) 25 r.p.m. (D) 75 r.p.m.
Ans: D
Ans: B
Over excitation gives leading power factor and under excitation gives lagging p.f .
Ans: D
Ans: A
The sheath in underground cable is provided to give mechanical strength.
Ans: D
Q.33 A 1:5 step-up transformer has 120V across the primary and 600 ohms resistance
across the secondary. Assuming 100% efficiency, the primary current equals
(A) 0.2 Amp. (B) 5 Amps.
(C) 10 Amps. (D) 20 Amps.
Ans: A
I1= V1 /R1 = 120/600 = 0.2 (η = 100%, losses are zero ∴V1 = VR = I1R1)
Q.34 A dc shunt generator has a speed of 800 rpm when delivering 20 A to the load at the
terminal voltage of 220V. If the same machine is run as a motor it takes a line
current of 20A from 220V supply. The speed of the machine as a motor will be
(A) 800 rpm. (B) more than 800 rpm.
(C) less than 800 rpm. (D) both higher or lower than 800 rpm.
Ans: C
Ng= Eg (60A / Φpz) Eg = V + Ia Ra ; in generator
Nm= Eb (60A / Φpz) Eb = V - Ia Ra ; in motor
Eg > E b for same terminal voltage
Therefore, Ng > N m
Q.35 A 50 Hz, 3-phase induction motor has a full load speed of 1440 r.p.m. The number of
poles of the motor are
(A) 4. (B) 6.
(C) 12. (D) 8.
Ans: A
N= Ns (1-S) = NS –NS x S
1440 = Ns (1-S)
Ns = 1440 / (1-S)
Ns = (120 f/ p) = 120 x 50/p = 6000 p
Ns will be closer to N i.e 1440
When P=2 ; Ns = 3000 rpm , not close to N
When P=4 ; Ns = 1500 rpm , it is closer to N
Therefore P =4 for N=1440
Ans: B
To make single phase motor self start. We split the phases at 90 degree. Hence, motor
behaves like a two phase motor.
Q.38 A synchro has
(A) a 3-phase winding on rotor and a single-phase winding on stator.
(B) a 3-phase winding on stator and a commutator winding on rotor.
(C) a 3-phase winding on stator and a single-phase winding on rotor.
(D) a single-phase winding on stator and a commutator winding on rotor.
Ans: C
Synchros : The basic synchro unit called a synchro transmitter. It’s construction
similar to that of a Three phase alternator.
Q.39 As the voltage of transmission increases, the volume of conductor
(A) increases. (B) does not change.
(C) decreases. (D) increases proportionately.
Ans: C
Decreases due to skin effect.
Q.44 Slip of the induction machine is 0.02 and the stator supply frequency is 50 Hz.
What will be the frequency of the rotor induced emf?
(A) 10 Hz. (B) 50 Hz.
(C) 1 Hz. (D) 2500 Hz.
10
Ans: C
Given : s = 0.02 ; f = 50 Hz
Therefore, frequency of rotor induced emf = s f
= 0.02 x 50 = 1.0 Hz
Q.45 A 4 pole lap wound dc shunt motor rotates at the speed of 1500 rpm, has a flux of 0.4
mWb and the total number of conductors are 1000. What is the value of emf?
(A) 100 Volts. (B) 0.1 Volts.
(C) 1 Volts. (D) 10 Volts.
Ans: D
Given N = 1500 rpm, Φ = 0.4 mWb, Z = 1000, P = 4, & A= 4
Therefore, Eb = NΦPZ / 60 A
= 1500 x 0.4 x 4 x 1000 x 10-3 / 60 x 4
= 60/6 = 10 volts
Q.46 The synchronous reactance of the synchronous machine is ______________.
(A) Ratio between open circuit voltage and short circuit current at constant field
current
(B) Ratio between short circuit voltage and open circuit current at constant field
current
(C) Ratio between open circuit voltage and short circuit current at different field
current
(D) Ratio between short circuit voltage and open circuit current at different field
current
Ans. A
The Synchronous reactance of a synchronous machine is a total steady state reactance,
presented to applied voltage, when rotor is running synchronously without excitation.
Therefore , XS = Ef / IS
= Emf of OC for same If / short circuit current
Q.47 A 3 stack stepper motor with 12 numbers of rotor teeth has a step angle of
____________.
(A) 12° (B) 8°
(C) 24° (D) 10°
Ans. D
Given m = 3, Nr = 12
Step angle = 360 / m x Nr = 360 /3 x 12 = 10°
11
Q.50 Inverse definite minimum time lag relay is also called ___________
(A) pilot relay. (B) differential relay.
(C) over current relay. (D) directional overcurrent relay.
Ans: B
Inverse definite minimum time lag relay characteristic is inverse but minimum time is
fixed. The operating time is inversely proportional to the magnitude of actuating
quantity.
12
Ans: D
Polarity test is required for parallel operation of transformers to know the direction of
current flow in secondary circuit w.r.t primary circuit.
Q.53 The short-circuit ratio of a typical synchronous machine is obtained from the OCC
and SCC curves of Fig.2 as
oa
(A)
ob
oa ′
(B)
ob ′
oa
(C)
ob ′
oc ′
(D)
ob
Ans: B
As shown in SCC curve the ratio of two field currents
Ans: A
Universal motor has same characteristics as DC series motor and also known as an a.c
series motor.
Q. 55 The rotor frequency for a 3 phase 1000 RPM 6 pole induction motor with a slip of
0.04 is________Hz
(A) 8 (B) 4
(C) 6 (D) 2
Ans: D
Given: N=1000 rpm ; P= 6; s= 0.04;
and f = N P/ 120
= 1000 6/120
= 50 Hz
Rotor frequency fr=s f = 0.04 50
= 2.0 Hz
13
Ans: C
If torque is zero then speed may exceed up to infinite, that is dangerous for machine
and machine can be damaged.
N 1/ T
Q.57 In the heating process of the ________type a simple method of temperature control is
possible by means of a special alloy which loses its magnetic properties at a particular
high temperature and regains them when cooled to a temperature below this value.
(A) Indirect induction over (B) core type induction furnace
(C) coreless induction furnace (D) high frequency eddy current
Ans: D
Magnetic property of alloy changes with change of the temperature and
Heat is produced due to eddy current = i2R and i f2
Q.58 In order to reduce the harmful effects of harmonics on the A.C. side of a high voltage
D.C. transmission system ______are provided.
(A) synchronous condensers (B) shunt capacitors
(C) shunt filters (D) static compensators
Ans: C
Xc= 1/ ωc
Q.59 An a.c. tachometer is just a ________with one phase excited from the carrier
frequency.
(A) two-phase A.C. servomotor (B) two-phase induction motor
(C) A.C. operated universal motor (D) hybrid stepper motor.
Ans: D
This is a special purpose machine whose stator coil can be energized by electronically
switched current.
14
Q.60 The torque, in a _____________is proportional to the square of the armature current
(A) DC shunt motor (B) stepper motor
(C) 2-phase servomotor (D) DC series motor
Ans: D
Ta .Ia and Ia ; therefore Ta Ia2
Q.61 The synchronous speed for a 3 phase 6-pole induction motor is 1200 rpm. If the
number of poles is now reduced to 4 with the frequency remaining constant, the rotor
speed with a slip of 5% will be _________.
(A) 1690 rpm (B) 1750 rpm
(C) 1500 rpm (D) 1710 rpm
Ans: D
Given : Ns1 =1200 , P1= 6,
P2 = 4, s = 0.05,
Frequency f = Ns P/120
= 120 6/120 = 60 Hz
rotor frequency f/ = s.f = 0.05 60 = 3.0 Hz
Now, Ns2 = 120 60 /4 = 1800 and Ns – N = 120 f / P2
Therefore, N=Ns- 120 f / P2 = 1800-120 0.05 60/4 = 1800-90 = 1710
Q.62 The eddy current loss in an a-c electric motor is 100 watts at 50 Hz. Its loss at 100
Hz will be
(A) 25 watts (B) 59 watts
(C) 100 watts (D) 400 watts
Ans: D
Eddy current losses f2
New loss (2f)2
New loss 4f2
∴ 4 times
Q.63 The maximum power for a given excitation in a synchronous motor is developed
when the power angle is equal to
(A) 0o (B) 45o
(C) 60o (D) 90o
Ans: A
P = VI cosΦ
Pmax = VI
∴Φ = 00
15
16
Q.69 A 3-phase, 400 votts, 50 Hz, 100 KW, 4 pole squirrel cage induction motor with a
rated slip of 2% will have a rotor speed of
(A) 1500 rpm (B) 1470 rpm
(C) 1530 rpm (D) 1570 rpm
Ans: B
N = NS (1-S) and NS =120 f / p
=120 x 50 /4 = 1500 rpm
∴N= 1500 (1-0.02) =1470 rpm
Q.70 If the phase angle of the voltage coil of a directional relay is 50 , the maximum
torque angle of the relay is
(A) 130 (B) 100
(C) 50 (D) 25
Ans: C
Torque Power
Power Voltage
Therefore, It has same angle as ‘V’ has.
Q.71 The voltage at the two ends of a transmission line are 132 KV and its reactance is
40 ohm. The Capacity of the line is
(A) 435.6 MW (B) 217.8 MW
(C) 251.5 MW (D) 500 MW
Ans: A
Line capacity is determined by power of line
P = (V2/R) or (V2/Z) when cos Φ =1
Q.72 A 220/440 V, 50 Hz, 5 KVA, single phase transformer operates on 220V, 40Hz
supply with secondary winding open circuited. Then
(C) Eddy current loss remains the same but hysteresis loss increases.
(D) Eddy current loss increases but hysteresis loss remains the same.
Ans: A
Wh = khfBm1.6 and We = kef2Bm2.k
Therefore, hysteresis and eddy current losses will be decreased when frequency
decreases.
17
Q.73 A synchronous motor is operating on no-load at unity power factor. If the field
current is increased, power factor will become
(A) Leading & current will decrease
(B) Lagging & current will increase.
(C) Lagging & current will decrease.
(D) Leading & current will increase.
Ans: A
Initially synchronous motor is operating at no load and unity power factor. When
field current increases, the excitation will increase. Therefore, p.f will be leading and
current will be I CosΦ < I
Q.74 A d.c. shunt motor runs at no load speed of 1140 r.p.m. At full load, armature reaction
weakens the main flux by 5% whereas the armature circuit voltage drops by 10%. The
motor full load speed in r.p.m. is
(A) 1080 (B) 1203
(C) 1000 (D) 1200
Ans: A
N2 / N1 =Eb2 /Eb1 x Φ 1 / Φ2 ; Φ2 = 0.95Φ 1 ; Eb2 = 0.9Eb1
∴ N2 /1140 = 0.9 x 1/0.95
N2 = 1080
Q.75 The introduction of interpoles in between the main pole improves the performance of
d.c. machines, because
(A) The interpole produces additional flux to augment the developed torque.
(B) The flux waveform is improved with reduction in harmonics.
(C) The inequality of air flux on the top and bottom halves of armature is
removed.
(D) A counter e.m.f. is induced in the coil undergoing commutation.
Ans: D
Counter e.m.f is produced, it neutralizes the reactive emf.
Q.76 The rotor power output of a 3-phase induction motor is 15 KW and corresponding slip
is 4%. The rotor copper loss will be
(A) 600 W. (B) 625 W
(C) 650 W (D) 700 W
18
Ans: B
Rotor copper losses = rotor input- rotor output
and output = (1-s) input
∴ Input = output/(1-s) = 15000 /1-0.04 = 15625
∴ loss = 15625 -1500 = 625 watt
Q.77 The direction of rotation of hysteresis motor is reversed by
(A) Shift shaded pole with respect to main pole
(B) Reversing supply lead
(C) Either A or B
(D) Neither A nor B
Ans: A
This motor used single phase, 50Hz supply and stator has two windings. These are
connected continuously from starting to running.
Q.78 A 1.8°step, 4-phase stepper motor has a total of 40 teeth on 8 pole of stator. The
number of rotor teeth for their rotor will be
(A) 40 (B) 50
(C) 100 (D) 80
Ans: B
Step angle ‘ ’ = NS – Nr / NS Nr x 3600
∴ 1-8 = -40 + Nr/40 Nr x 3600
Nr = 50
Q.79 Low head plants generally use
(A) Pelton Turbines (B) Francis Turbine
(C) Pelton or Francis Turbine (D) Kaplan Turbines
Ans: A
In the hysterisis motor, the direction of rotation can be reversed by shifting the
shaded pole region with respect to main pole. But not by changing supply lead
because it has ac supply.
Q.80 The charging reactance of 50 Km length of line is 1500 . The charging reactance for
100Km length of line will be
(A) 1500 (B) 3000
(C) 750 (D) 600
19
Ans: B
Characteristic reactance per km = 1500/50 = 30 ohms
∴ Characteristic reactance per 100km = 30 x 100 = 3000 ohms
Q.81 Electric ovens using heating elements of _______________ can produce temperature
upto 3000°C.
(A) Nickel (B) Graphite
(C) Chromium (D) Iron
Ans: C
Chromium has high melting point.
Ans: A
Transformers having higher kVA rating will share more load.
Q.84 As compared to shunt and compound DC motors, the series DC motor will have the
highest torque because of its comparatively ____________ at the start.
(A) Lower armature resistance. (B) Stronger series field.
(C) Fewer series turns. (D) Larger armature current.
Ans: D
T Φ Ia (before saturation)
Φ Ia
T Ia 2
Q.85 A 400kW, 3-phase, 440V, 50Hz induction motor has a speed of 950 r.p.m. on full-
load. The machine has 6 poles. The slip of the machine will be _______________.
(A) 0.06 (B) 0.10
(C) 0.04 (D) 0.05
20
Ans: D
N = Ns (1-S)
950 = 120 x 50 (1-S)/6
S = 0.05
Q.86 Reduction in the capacitance of a capacitor-start motor, results in reduced
(A) Noise. (B) Speed.
(C) Starting torque. (D) Armature reaction.
Ans: C
Reduction in the capacitance reduces starting voltage, which results in reduced
starting torque.
Q.87 Regenerative braking
(A) Can be used for stopping a motor.
(B) Cannot be easily applied to DC series motors.
(C) Can be easily applied to DC shunt motors
(D) Cannot be used when motor load has overhauling characteristics.
Ans: B
Because reversal of Ia would also mean reversal of field and hence of Eb
Q.88 At present level of technology, which of the following method of generating electric
power from sea is most advantageous?
(A) Tidal power. (B) Ocean thermal energy conversion
(C) Ocean currents. (D) Wave power.
Ans: A
At present level of technology, tidal power for generating electric power from sea
is most advantageous because of constant availability of tidal power.
Q.89 If the field circuits of an unloaded salient pole synchronous motor gets suddenly open
circuited, then
(A) The motor stops.
(B) It continues to run at the same speed.
(C) Its runs at the slower speed.
(D) It runs at a very high speed.
Ans: B
The motor continues to run at the same speed because synchronous motor speed
does not depend upon load, Nα f.
21
Ans: B
Disc type electrodes are used for electric resistance seam welding.
Q.91 For LV applications (below 1 kV), ______________ cables are used.
(A) Paper insulated. (B) Plastic.
(C) Single core cables. (D) Oil filled.
Ans: C
For low voltage applications single core cables are suitable.
Ans: B
The primary input current under no load conditions has to supply (i) iron losses in
the core i.e hysteresis loss and eddy current loss (ii) a very small amount of Cu loss
in the primary (there being no Cu loss in secondary as it is open)
Q.93 A transformer operates most efficiently at 3/4th full load. Its iron (PI) and copper
loss (PCu) are related as:
(A) PI PCu = 16 9 (B) PI PCu = 4 3
Ans: D
If PCu is the Cu loss at full load, its value at 75% of full load is
PCu x (0.75)2 = 9/16 PCu
At maximum efficiency, it equals the iron loss PI which remains constant through
out. Hence max. efficiency at
22
PI = 9/16 PCu
Or PI / PCu = 9/16
Ans: C
Since reluctance on the q axis is higher, owing to the larger air gap, hence xq < xd
Ans: A
Thinner the laminations, greater is the resistance offered to the induced e.m.f.,
smaller the current and hence lesser the I2R loss in the core.
Ans: A
Mechanical Power developed by the rotor (Pm) or gross power developed by rotor
(Pg)
= rotor input –rotor Cu losses
= (3I/2 R2/ / S) -(3I/2 R2/ )
= 3I/2 R2/ (1/ S -1)
23
Ans: A
The rotor revolves synchronously because the rotor poles magnetically lock up with
the revolving stator poles of opposite polarity
Ans: D
Temperature of resistance furnaces can be controlled by changing either applied
voltage or by number of heating elements or by circuit configuration.
Ans: B
The line trap unit employed in carrier current relaying offers high impedance to
carrier frequency signal.
Because carrier frequency range is 35 km – 500 kHz
XL = 2 f l
Where f increases XL will also increases
24
NUMERICALS
Q.1 Calculate the voltage regulation of a transformer in which ohmic drop is 2% and
the reactance drop in 5% of the voltage at 0.8 lagging power factor. (7)
Ans: The expression for % voltage regulation is
V 2 0 − V 2 , fl
% voltage regulation= X1 0 0 …(1)
V 2 , fl
where V2 fl = rated secondary voltage while supplying full load at a specified power
factor.
I ( R cos φ + X sin φ )
% Voltage regulation = X 100
V2
IR IX
= .100 cos φ + .100 sin φ _ …(2)
V2 V2
The quantities within the brackets are given in the problem as 2%( percent ohmic
drop) and 5% (percent reactance drop). Also φ is the lagging power factor angle. The
plus sign in Equation (2) is because of the lagging nature of current.
Now
= (1.6 + 3.0)%
= 4.6%
25
Fig. C1 shows the flux density wave in the air gap and the conductor current
distribution in the developed armature for one pole-pair. The force on the
conductors is unidirectional. Each conductor, as it moves around with the
armature, experiences a force whose time variation is a replica of the flux
density(B).
Therefore, the average conductor force
f c ( av ) = Bav l I c (1)
where Bav = average flux density over a pole.
l = active conductor length, and I c =conductor current
Total force
F = zf c ( av ) = Bav I c lz , where z=total number of conductors (2)
This force (and therefore torque) is constant because both the flux density wave and
current distribution are fixed in space at all times. Now the torque developed is
T = Bav I c lzr _ (3)
where r is the mean air gap radius
The flux/pole can be expressed as
φ = Bavτ p l (4)
2π r
where τ p = polepitch =
P
26
so
2π r
φ = B av l
P
or
φP 1
Bav = X (5)
2π rl
zP
T = K aφ I a _ where K a = =constant (6)
2π A
Q.3 A 230 V d.c. shunt motor with constant field drives a load whose torque is
proportional to the speed. When running at 750 rpm it takes 30 A. Find the speed
at which it will run if a 10 ohm resistance is connected in series with the armature.
The armature resistance may be neglected. (7)
Ra
230 V
Ia
27
T= K aφ I a
T1 K a φ1 I a 1 Ia
Now = = 1 (1)
T2 K aφ 2 I a 2 I a2
T1 N1 I a1
= = from equation (1) (2)
T2 N 2 I a 2
N2
And I a 2 = I a1 (3)
N1
Ea1 VT − 0 K φN
Back emf = = a 1
Ea2 VT − I a2 R Kaφ N2
230 − I a 2 X 10 N 2
Or =
230 N1
230 − 10 X I a1 X
N2
N1 N2
Or =
230 N1
230 −10 X 30 X
N2
N1 N2
Or =
230 N1
This gives
N2
( 230 + 300) = 230
N1
28
Therefore
230
N 2 = N1
530
230
= 750 X
530
= 326rpm
Q.4 The power input to a 500 V, 50 Hz, 6 pole 3 phase squirrel cage induction motor
running at 975 rpm is 40 KW. The stator losses are 1 KW and the friction and
windage losses are 2 KW. Calculate
120 f 50
Ans: Synchronous speed(N s )= = 120 X = 1000RPM
P 6
ωs − ωm 2π X 1000 − 2π X 975 25
Slip = = = = 0.025
ωs 2π X 1000 1000
Power across air gap(PG) = power input - stator copper loss
Thus PG=40KW-1KW=39KW
= (38.025-2.000)KW
=36.025KW
Net Mech output 36.025
Efficiency = =
Power input 40
≈ 90%
Note: Assume that the core loss is included in friction and windage loss and the total
loss under this head is 2.0 kW
29
Q.5 A 120V, 60Hz, 1/4hp universal motor runs at 2000 rpm and takes 0.6A when
connected to a 120V d.c. source. Determine the speed, torque and power factor of the
motor when it is connected to a 120V, 60 Hz, supply and is loaded to take 0.6A (rms)
of current. The resistance and inductance measured at the terminals of the motor are
20 ohm and 0.25H respectively. (7)
Ans: Universal Motor: (A) When connected to a d.c. source it runs at 2000RPM and
takes 0.6A [Fig F1]
0.6A
20Ω
120V
DC Source Eb
1 φ n 2 Z P Kn ac
E a ( ac ) RM S = * =
2 60 A 2
Rmotor=20 Ω
Lmotor=0.25H
Xmotor=2 π X60X0.25
Or, Xmotor=94.25 Ω
From the phasor diagram
Ea( ac ) + Ia R = V 2 − (Ia X )2
30
Assume, same flux for the same current (i.e. 0.6 Adc and 0 .6 Arm s )
Therefore
93.84
nac = 2000 X = 1737.78rpm
108
Ea( ac ) + Ia R 93.84 +12
Power factor, cosφ = = = 0.88 lag
V 120
Pmech = Ea( ac ) I a = 93.84 X 0.6 = 56.3W
Pmech 56.3
Tdev = = = 0.309N − m
ωm 2π X
1737.78
60
Q.6 For a 4 KVA, 200/400 V, 50 Hz, 1 – phase transformer, calculate the efficiency,
voltage at the secondary terminals and primary input current when supplying a full
– load secondary current at 0.8 lagging power factor.
The following are the test results:
Open circuit with 200 V applied to the L.V. side: 0.8 A, 70 W. Short circuit with
20 V applied to the H.V. side: 10 A, 60 W. (14)
Ans: The transformer is supplying full-load secondary current at 0.8 lagging power
factor
4 KVA 4000VA
Full load secondary current = = = 10 A
400V 400V
From the open circuit test, core losses = 70W
From the S.C. test, full load copper losses = 60W
(a) Efficiency
V2 I 2 cos θ
η= X 100
V2 I 2 cos θ + core losses + full load copper losses
31
4000 X 0.8
= X 100
4000 X 0.8 + 70 + 60
3200
= X 100 = 96.1%
3300
(b) voltage at the secondary terminals is determined as follows with the help of
equivalent circuit of Fig A3
Req Xeq
200V E1 E2
V1
I2
a X e/
θ E 1 = V2 a I2
a R e/
32
Q.7 Draw the per phase approximate equivalent circuit of a 3 – phase induction motor
at slip ‘s’ and derive the expression for electromagnetic torque developed by the
motor. Derive also the condition for maximum torque and the expression for the
maximum torque. (14)
Ans:
R1 X1 X’2
a
I1 I0 I’2
V
Xm R ’2
R1
a
Fig B3 shows the per-phase exact equivalent circuit of a 3-phase induction motor. The
power crossing the terminals ’ab’ in Fig B3 is the electrical power input per phase
minus the stator copper loss and iron loss; Thus it is the power that is transferred from
the stator to the rotor via the air gap. It is also known as the power across the air gap.
3(I 2′ ) R 2′
2
33
3 ( I 2′ ) R2′
2
PG
or T = =
ωs sωs
dT
The condition (that is, slip) for maximum torque is obtained by equating to zero.
ds
With the approximation of I1 = I 2′ in Fig. B3 or using the approximate equivalent
circuit.
V
I 2′ = 1/ 2
( R1 + R2′ / s ) + ( X 1 + X 2′ )
2
3( I2′ ) R2′
2
′ 2
3RV
T= = 2 _____________________________
(I)
sωs sωs ( R1 + R2′ / s) +( X1 + X2′ )
2 2 1/2
dT
Equating to zero gives the slip at maximum torque as
ds
R2′
sm = ±
2 1/ 2
R12 + ( X 1 + X 2′ )
3 V2
Tmax =
2ωs R + R 2 + X + X ′ 2
1 1 ( 1 2)
Q.8 A 230 V d.c. series motor has an armature resistance of 0.2 Ω and series field
resistance of 0.10 Ω . Determine:
(i) the current required to develop a torque of 70 Nm at 1200 rpm
(ii) percentage reduction in flux when the machine runs at 2000 rpm at half the
current. (14)
Ans:
DC Series Motor:
φ ZP n ZP
Back emf Ea = . = φn
60 A 60 A
60 A
Define K n =
ZP
34
φn
So Ea = _____________
( A)
Kn
1 P ZP
Also T = φ Ia Z = φ Ia .
2π A 2π A
ZP ZP 60 1 60
= . = .
2π A 60 A 2π K n 2π
φI 60
So T= aX
K n 2π
Also for a series motor
Ea = V-Ia (Ra+Rse)----------------(B)
where V is the applied voltage and Ia is the current through armature and series field.
nφ
= 230 − I a (0.2 + 0.1)
Kn
1200φ
= 230 − 0.3I a _____________ (C )
Kn
60 φ
also 70 = .I a
2π K n
or
φ 70 2π
= X
_____________
( D)
Kn 60 I a
1200 X 70 X 2π
= 230 − 0.3I a
60 I a
8796
or = 230 − 0.3I a
Ia
35
n2φ2
= 230 − 0.3 X 20.17
Kn
φ2
2000 = 230 − 6.05
Kn
φ2
2000 = 224 ______________ ( E )
Kn
1200φ
From (C) = 230 − 0.3(40.33) = 217.9 _____________ ( F )
Kn
Ans: The occ and scc characteristics of the synchronous generator are given in
Fig. M4
36
1100
Thus Xs ≈ Zs = = 3.18Ω
3 X200
E f − Vt ( rated )
Percentage voltage regulation is defined as X100%
Vt ( rated )
The phasor diagram is given in Fig. M5
Ef
IXs
δ Vt(rated)
φ θ
cos φ = 0.8
Ia = 200A
37
2200
Here Vt ( rated ) =
3
cos φ = 0.8
E f 2 = (Vt + Ira cos φ + IX s sin φ ) + ( IX s cos φ − Ira sin φ ) 2
2
where Vt(terminal voltage) and Ef (field voltage) are per phase values
E f − Vt ( rated )
Percent regulation = X 100
Vt ( rated )
1788.8 − 1270
% reg n = X 100 = 48.85%
1270
448.8
tan δ = = 0.259
1731.6
δ =14.5
Thus power angle = 14.5
Q.10 A 100 KVA, 2400/240 V, 50 Hz, 1-phase transformer has no-load current of 0.64
A and a core loss of 700 W, when its high voltage side is energized at rated voltage
and frequency. Calculate the two components of no-load current. If this transformer
supplies a load current of 40 amp at 0.8 lagging power factor at its low voltage
side, determine the primary current and its power factor. Ignore leakage impedance
drop. (12)
Ans:
38
No load: -
Io = 0.64 A
Wo= 700W
Iron loss current = 700/2400 =0.2916 A
Now , Io2 = Iw2 + Iµ2
= (0.642 – 0.29162)1/2
= 0.5697 A
On load :-
I2 = 40 A
Φ2 = 0.8 lag
= 62.88o
Φ2 = cos –1 0.8
= 36.86o
Now, turn ratio K =V1/V2 = 240/ 2400 = 0.1
39
= 26.02o
I0 = 0.64 ∠–62.88o
I21 = 4∠–36.86o
Ans:
Eg = 254 V
V = 240 V
Ra= 0.04 Ω, Rsh = 24Ω
Eg = V + Ia Ra
Eg = V + (IL + Ish) Ra
Ish = V / R sh = 240/24 = 10A.
Substituting the values in the above expression,
254 = 240 + (IL + 10) 0.04
IL = 340A
40
Q.12 The shaft output of a three-phase 60- Hz induction motor is 80 KW. The friction
and windage losses are 920 W, the stator core loss is 4300 W and the stator copper
loss is 2690 W. The rotor current and rotor resistance referred to stator are
respectively 110 A and 0.15 Ω . If the slip is 3.8%, what is the percent efficiency?
(12)
Ans:
Pm =output = 80 KW
41
Ans:
N = 950 rpm
V= 550 V
f = 50 Hz.
cosϕ = 0.88
= 32.34 ampere
42
Ans:
P=120 * 50/1000=6;
Q.15 Three single-phase, 50 kVA, 2300/ 230 V, 60 Hz transformers are connected to form
a 3-phase, 4000V / 230-V transformer bank. The equivalent impedance of each
transformer referred to low-voltage is 0.012 + j 0.016 Ω . The 3-phase transformer
supplies a 3-phase, 120 kVA, 230 V, 0.85 power-factor (lagging) load.
Ans: Given :
43
44
Q.16 A pair of synchronous machines, on the same shaft, may be used to generate power at
60 Hz from the given source of power at 50 Hz. Determine the minimum number of
poles that the individual machines could have for this type of operation and find the
shaft-speed in r.p.m. (4+4)
Ans: Motor & generator (synchronous machine) are coupled. Therefore,
NS(m) = NS(g)
NS(m) = 120 fm /Pm ; NS(g) = 120 fg / Pg
Where : NS(m) = synchronous speed of motor
NS(g) = synchronous speed of generator
fm = frequency of motor power
fg = frequency of generator power
Pm = motor poles
Pg = generator poles
∴ 120 fm /Pm = 120 fg / Pg
120 x 50 / Pm = 120 x 60 /Pg
∴ Pg /Pm = 6/5
Pg : Pm = 6 : 5
Therefore minimum requirement of poles for motor
Pm = 10 (5 x 2)
Pg = 12 (6 x 2)
Now synchronous speed or shaft speed = 1200 x 50 / 10 = 600 rpm
Q.17 A 240V dc shunt motor has an armature resistance of 0.4 ohm and is running at the
full-load speed of 600 r.p.m. with a full load current of 25A. The field current is
constant; also a resistance of 1 ohm is added in series with the armature. Find the
speed (i) at the full-load torque and (ii) at twice the full-load torque.
(6)
Ans: In a DC shunt motor
V = 240V
Ra = 0.4
N1 = 600rpm (full load speed)
Ia = 25A and, ISh is constant
R =1 added in series with armature
Eb1 = V - IaRa
45
Q.18 A 400V, 4-pole, 50 Hz, 3-phase, 10 hp, star connected induction motor has a no
load slip of 1% and full load slip of 4%. Find the following:
(i) Syn. speed (ii) no-load speed (iii) full-load speed.
(iv) frequency of rotor current at full-load (v) full-load torque. (5 x 2 = 10)
Ans: Given :
VL = 400 volts ; P = 4 nos, 50 Hz,
P0 = 10 HP = 735.5 x 10 = 7355 watt
i. Synchronous speed NS = 120 f / p
= 120 x 50 / 4 = 1500 rpm
ii. No load speed at s = 0.01
N0 = NS ( 1 – s) = 1500 ( 1- 0.01)
= 1485 rpm
iii. Full load speed at sf = 0.04
Nfl = NS (1-sf ) = 1500 ( 1-0.04)
= 1440 rpm
46
47
48
49
50
Q.25 A 250 Volts dc shunt motor has Rf=150 ohm and Ra = 0.6 ohm. The motor
operates on no-load with a full field flux at its base speed of 1000 rpm with Ia = 5
Amps. If the machine drives a load requiring a torque of 100 Nm, calculate
armature current and speed of the motor. (8)
Ans: Given V = 250 V
Rf = 150 Ω
Ra = 0.6Ω
i. operate no load and full load flux
No = 1000 rpm & Iao = 5 amp.
51
Q.26 A 400Volts, 1450 rpm, 50 Hz, wound-rotor induction motor has the following
circuit model parameters.
R1= 0.3 ohm R2=0.25 ohm
X1=X2=0.6 ohm Xm= 35 ohm
Rotational loss =1500 W. Calculate the starting torque and current when the motor
is started direct on full voltage. (8)
52
Ans:
V = 400V; N = 1450 rpm ; f =50 Hz
R1 = 0.3 Ω R2’ = 0.25 Ω ; X1 =X2’ = 0.6 Ω ; X0 = 35Ω
Rotational losses =1500 W
Find starting torque T1 = ? If =?
Or If = V/Z01 = 400/√3
1.32
= 175.16 amp.
53
(N = (1-S)NS )
= 606.86 - 1500/151.67
= 596.97 Nm
Q.27 A universal motor (ac–operated) has a 2-pole armature with 960 conductors. At a
certain load the motor speed is 5000 rpm and the armature current is 4.6 Amps, the
armature terminal voltage and input power are respectively 100 Volts and 300
Watts.
Compute the following, assuming an armature resistance of 3.5 ohm.
(i) Effective armature reactance.
(ii) Maximum value of useful flux/pole. (8)
Ans: P = 2; Z = 960; N= 5000
Ii =Ia= 4.6 amp ; V1 =100 volts
Pi =300W find Xa and φm = ?
P1 = V1I1cos φ
∴ cos φ = P1 / V1I1
= 300/ 100 x 4.6 = 0.652
Ebdc = V - IaRa or (NφPZ / 60A )
= 100 - 4.6 x 3.5
= 100 -16.1 = 83.9 volts
54
Xa = 16.48 ohms
∴ φm = E bdc x 60 A/ NPZ
Flux
= 1.048 x 10-3 wb
Ans:
Given 1 φ, 50 Hz ; cosφ = 0.707 lagging
Transmission length = 20 km
Generator supply inductive load = 5000kW = kVAR
R = 0.0195 ohms/km , L = 0.63 mH
VR = 10kV
Find sending end voltage VS = ? & % regulation =?; distance =20km
∆V = VS -VR ( drop in line )
(VS -VR )=RP + XQ/ VR {Active Power (P) = Reactive Power (Q)}
= 0.0195 x 20 x 5000 x 103 + 3.96 x 5000 x 10-3 / 10 x 103
Vs ={0.0195 x 20 x 5000 x 103 + 3.96 x 5000 x 10-3 / 10 x 103 } + 10 x
103 volts
= 1.2175 x 104 volts
= 12.175 kVolts
% Regulation = (VS -VR )/ VS x 100
= 17.86%
55
Q.29 A 37.7 HP, 220 V d.c shunt motor with a full load speed of 535 rpm is to be braked
by plugging. Estimate the value of resistance which should be placed in series with
it to limit the initial current to 200 A. (8)
Ans:
P0 = 37.7 HP = 37.7 x 735.5 W = 27.73 kW
NL = 535 rpm braked by plugging ; Ia = 200A
V =220 v
Under plugging :
Ia = V + Eb / (R + Ra )
and Eb = 27.73 x 103 / 200 (assume negligible losses)
= 138.65 Volts
200 = 220 + 138.65/(R + Ra )
R + Ra = 358.65 /200 = 1.79 ohm
R = value of added resistance in series with armature resistance
Ra = armature resistance
56
= 97.28 %
(ii) At half load and unity p.f.
Wc = 480 ( ½)2 = 120 W
Wi = 360 W
Total losses = 360+120 = 480 W
At half load, output at unity p.f = 30/2 1= 15 kW
Therefore, Efficiency = (15/(15+0.48)) 100
= 96.90 %
(iii) The load for maximum efficiency and condition for max. efficiency.
Efficiency ( ) = ( V1I1 cos - losses )/ V1I1 cos
= ( V1I1 cos - I12R01- Wi )/ V1I1 cos
I 1 Ro 1 Wi
=1 −
V 1 R 1 cos V 1 R 1 cos
Differentiating above equation for maximum efficiency
Therefore,
dn R01 Wi
=0- - =0
dI1 V1cos V1I1cos
and
R01 Wi
=
V1cos V1I1cos
Therefore,
Wi = Wc
and load current, Il = ( Wi /R02 )
Ans:
Given : VL = 33 kV, 3 phase, star connected Synchronous Generator
∴ Vph = 33/ 3 = 19.1 kV
Zph = 1.4
P01 = 240 MW = 240 x 106 Watt
cos Φ 1 = 1
2 = 1.25 1 (Flux / excitation)
P02 = 280 MW = 280 x 106 Watt
Find IL2 = ? ; cos Φ 2 =?
57
Ψ2
Eg2 /ph = Eg1 /ph *
Ψ1
∴ Eg2 /ph = 31.23kV
or Eg2 /ph = Vph + I2 Zph
I2 ph = Eg2 -V/Zph = 12.125 x 103 / 1.4
= 8.66 kA
Now P02 = 3 VL IL2 cos Φ 2 ( IL2 = Iph 2 : VL = VL1 = VL2 ; P02 = 280 MW )
cos Φ 2 = P02 / 3 VL IL2
= 280 x 10 6 / 3 x 33 x 8.66 x 106
= 280 / 495
= 0.565 leading
Q.32 A 250 V DC shunt motor has an armature resistance of 0.55 Ω and runs with a full
load armature current of 30A. The field current remaining constant, if an additional
resistance of 0.75 Ω is added in series with the armature, the motor attains a speed of
633 rpm. If now the armature resistance is restored back to 0.55 Ω , find the speed
with (i) full load and (ii) twice full load torque. (12)
Ans:
Given Ra =0.55 ,
RT =0.75 additional resistance in series with armature
V = 250V
Ia = 30A at full load Then N2 = 633 rpm
Ish = Constt.
Find N1 = ? at full load ; N3 = ? at double load
N1 V - Ia1Ra
58
N2
= 1.106
N1
N1 = N2 *1.106 = 700.5
N1 = 701 r.p.m at full load without Rt
N 1 V − I a1 Ra
=
N 3 V − I a3 Ra
N1
= 1 . 076
N3
Q.33 A 4-pole, 3 phase, 400 V, 50 Hz, induction motor has the following parameters for its
circuit model (rotor quantities referred to the stator side) on an equivalent-star basis:
R 1 = 1.6 . X1 = 2.4 . R 12 = 0.48 Ω, X 12 = 1.2Ω and X m = 40 . Rotational
losses are 720 W. Neglect stator copper losses. For a speed of 1470 rpm, calculate the
input current, input power factor, net mechanical power output, torque and efficiency.
(12)
Ans:
Given:
P =4
VL = 400 V
f = 50 HZ
R1 = 1.6 , X1 = 2.4 ; R2/ =0.48 ; X2/ = 1.2
Xm = X0 = 40
Rotational losses = 720 W; Wi = 0 Watt
N=1470 rpm
Find I1 = ? ; cos = ? ; Pmo = ?; TS = ? ; η=?
NS = 120 f / P = 120 x 50 / 4 = 1500 rpm
59
60
Q.34 A universal motor has a 2-pole armature with 1020 conductors. When it is operated
on load with a.c. supply with an armature voltage of 150, the motor speed is 5400
RPM. The other data is:
Input power : 360 W
Armature current : 5.2 A
Armature resistance: 5.5
Compute (i) the effective armature reactance and (ii) maximum value of armature
flux per pole. (10)
Ans: Given:
P=2; Va = 150 V; N = 5400 rpm
Z=1020; Pi = 360 MW; Ia = 5.2A;
Ra = 5.5
Find Xa =?; Φmax per pole=?
61
Ans:
Given : P0 = 50 kVA, V1 = 2300V , V2 = 230V, f = 60 Hz
R1 = 0.65 (H.V.Side), R2 = 0.0065 (L.V Side)
Lab test: O.C (H.V.Side) : V= 230 V, I =5.7A, P = 190 watt
S.C (L.V.Side), : V= 41.5 V, I =21.7A, P = ? watt
Find (a) V1 for rated V2 when acts as step up transformer and delivering 50kVA at
cos Φ = 0.8
(b) efficiency
At S.C test
Z02 = VSC / I2 = 41.5 /21.7 = 1.912 & K =1/10
∴Z01 = Z02 / K2 = 1.912 x 102 = 191.2
R01 = R1 + R2/ = 0.65 + 0.0065/100 (K =1/10, H.V side is R2)
R01 = 0.13
R02 = R01 x K2 = 0.13 / 100 = 0.0013
Now X01 = Z012 – R201 = (191.2)2 – (0.13)2
X01 = 191.2
& X02 = Z022 – R202 = (1.912)2 – (0.0013)2
X01 = 1.912
Total transformer voltage drop referred to secondary
VD2 = I2 ( R02 cosΦ2 + X02 sinΦ2)
= 21.7 ( 0.0013 x 0.8 + 1.92x 0.6) = 21.7 (1.153)
= 25.02 Volts
∴V1/ = V1 + VD2 = 230 + 25.02 = 255.02 Volts.
62
∴V1 =255.02 volts for V2 = 2300 volts rated value as step up.
= 50 x 1000 x 0.8 .
output + losses (core Losses + Cu losess)
= 40000 .
40000 + 190 + I1 R1+ I22 R2
2
= 40000 .
40000 + 190 + (5.7)2 x 0.65+ (21.7)2 x 0.0065
(note: low voltage winding is short circuited)
= 40000 .
40000 + 32.49 x 0.65+ 470.89 x 0.0065
= 40000 .
40000 + 21.12 + 3.06
= 99.93%
x φ = 77Ω
The rotational losses are 4100 watts. Using the approximate equivalent circuit,
compute for a slip of 1.5%.
a. the line power factor and current.
b. developed torque.
c. efficiency. (8+4+4)
Ans:
Given: 3 phase , 335 HP, 50 Hz, V = 2000V Induction motor
P0 = 335 x 735.5 watt Rotational losses = 4100 watt; Slip (s) = 0.015
P = 6 nos., Y connected
∴ IL = Iph & VL = 3 Vph
Vph = VL / 3 = 2000 / 3 = 1154.70 volts
63
64
Q.37 A 2300-V, three phase, 60 Hz, star-connected cylindrical synchronous motor has a
synchronous reactance of 11 Ω per phase. When it delivers 200 hp, the efficiency is
found to be 90% exclusive of field loss, and the power-angle is 15 electrical degrees
as measured by a stroboscope. Neglect ohmic resistance and determine:
(a) the induced excitation per phase.
Ans:
Given : 2300V , 3 phase, 60Hz, Synch. Motor
XS = 11 /ph , Star connected, VPh = 2300/ 3 V
P0 = 200 hp = 200 x 735.5 = 147.1 kW
η = 90%
∴ Pi = P0 /0.9 = 163.44 kW
Power angle = 150 ( electrical)
Find: induced excitation / ph (Eg ) = ?
Line Current Ia = ?
Power Factor Cos Φ =?
65
Q.38 When a 250-V, 50 hp, 1000 rpm d.c shunt motor is used to supply rated output power
to a constant torque load, it draws an armature current of 160A. The armature circuit
has a resistance of 0.04 Ω and the rotational losses are equal to 2 KW. An external
resistance of 0.5 Ω is inserted in series with the armature winding. For this condition
compute
(i) the speed
(ii) the developed power
(iii) the efficiency assuming that the field loss is 1.6 K.W (8+4+4)
Ans:
Given VL = 250 V P0 = 50 hp = 50 x 735.5 36.78 kw
N1 = 1000 rpm Ia = 160 amp. , Ra= 0.04 , R = 0.5 ,
Rotational losses = 2 kw ; Field losses = 1.6 kW
66
= 17.42 kW
= 33.46%
Q.39 The following data were obtained on a 20KVA, 50Hz, 2000/200V distribution
transformer
Open Circuit Test (on L.V. side): 200V, 4A, 120W
Short Circuit Test (on H.V. side): 60V, 10A, 300W
Draw the approximate equivalent circuit of the transformer referred to H.V. Side. (8)
Ans:
Given : 20 kVA, 50Hz , 2000/ 200V
O.C Test : V0 = 200 V ; I0 = 4A ; W1 = 120 W
S.C Test : VSC= 60 V ; ISC = 10A ; WSC = 300 W
V0I0 cosΦ0 = W0
∴ cosΦ0 = W0 / V0I0 = 120 / 200 x 4 = 0.15
∴ sinΦ 0 = 0.988
67
Q.40 The efficiency of a 3-phase 400V, star connected synchronous motor is 95% and it
takes 24A at full load and unity power factor. What will be the induced e.m.f. and
total mechanical power developed at full load and 0.9 power factor leading? The
synchronous impedance per phase is (0.2+j2) . (9)
Ans:
Given : 3Φ , 400V star connected synchronous motor
Output = 95% of input
VPh = 400/ 3
at p.f = 1 ; Ia = 24 amp. ; V = VPh = 230.94 volt
ZS = (0.2 + j2) = 2 84.29
Find : Eb at 0.9 p.f leading
Mechanical power developed ?
68
cos Φ = 0.9
∴ Φ = 25.840
ER = IZS = 24 (0.2 +j2) volts
= (4.8 +j48) volts = 48.23 84.290
θ =84.290
Now at leading p.f
Eb/ph = V + IaZS cos {180 – (θ + Φ)} + j IaZS sin {180 – (θ - Φ )}
∴ Eb/ph = 231+24x2cos {180 – (84.29 + 25.84)} + j24x2 sin
{180 – (84.29 - 25.84)}
=231 + 48 cos (69.87) + j 48 x 0.938
= 231 + 16512 + j45
Eb/ph = 247.5 + j 45
Eb/ph = 251.55 10.3
Synchronous motor input power = 3 VLILcos Φ
= 3 VIacos Φ = 3 x 400 x 24 x 0.9
= 14,964.92 watt
total copper losses = 3 I2 Ra
= 3 x (24)2 x 0.2 = 3 x 576 x 0.2
= 345.6 watt
∴ Mechanical output developed = Input – losses
= 14964.92 -345.6
= 14619.32 watt
Q.41 A 200V shunt motor with a constant main field drives a load, the torque of which
varies at square of the speed, when running at 600 r.p.m., it takes 30A. Find the speed
at which it will run and the current it will draw, if a 20 resistor is connected in series
with armature. Neglect motor losses. (9)
Ans:
Given: V =200v, shunt motor
N1 = 600 rpm
I1 = 30A = Ia1
Find : N2 & I2 ; when R =20 added with Ra in series
69
Q.42 A 3-phase induction motor has a starting torque of 100% and a maximum torque of
200% of full load torque. Find
(i) Slip at maximum torque.
(ii) Full load slip.
(iii) Neglect the stator impedance (8)
let a = R2 / X2
2a /(1+a2) = 1/2
70
4a = 1+a2
a2 - 4a + 1 = 0
a=4± 16 – 4 x 1 x 1
2
a=4± 12
2
a = 3.73 or 0 .2679
Sf = 0.9995 or 0.07145
Q.43 A universal motor (a.c. operated) has a 2-pole armature with 960 conductors. At a
certain load the motor speed is 5000 r.p.m. and the armature current is 4.6A. The
armature terminal voltage and input are respectively 100 V and 300 W. Compute the
following, assuming an armature resistance of 3.5 .
(i) Effective armature reactance
(ii) Max. value of useful flux per pole. (8)
71
Ans: Given:
3Φ , 50Hz, d = 250km
P0 = 25mVA, cos Φ = 0.8 (lagging), VR = 132 KV
VR/ph = 132000/ 3 V
Spacing between conductors = 3m
R = 0.11 /km
Dia (D) = 1.6 x 10-2 m
Find : Vs= ? and % regulation =?
Iph = Po / VR/ph
I = 25 x 106/ 132 x 1000 = 25000/132
Line loss = 3 I2 R = 3 x (25000 / 132)2 x 0.11 x 250
Resistance / phase = 0.11 x 250 = 27.5
VS/ph = VR/ph + IZ ; ( Z = R/Phase)
72
For Y : Vp = VL/ 3 , Ip = IL
: Vp = VL , Ip = IL/ 3
VPP /VPS = N1 / N2
VLP /VLS = 3 VPP/VPS = 3 N1/N2 = 12 3
VLS = VLP / 12 3 = 6600 / 12 3
VLS = 315.33 Volts
IPP / IPS = N2 / N1
ILP / ILS = IPP / 3 x 1/ IPS =(1/ 3 ) x ( N2 / N1)
73
(ii) /Y
For Y : Vp = VL/ 3 , Ip = IL
: Vp = VL , Ip = IL/ 3
VPP /VPS = N1 / N2
VLP /VLS = VPP/ 3 VPS = 1/ 3 x N1/N2 = 12 / 3
VLS = 3VLP / 12 = 3 x 6600 / 12
VLS = 550 3 Volts
IPP / IPS = N2 / N1
ILP / ILS = 3IPP / IPS = 3 N2 / N1 = 3/12
Q.46 A 3300V, delta-connected motor has a synchronous reactance per phase (delta) of
18 . It operates at a leading power factor of 0.707 when drawing 800kW from the
mains. Calculate its excitation emf. (8)
74
75
Q.48 A 150kW, 3000V, 50Hz, 6-pole star-connected induction motor has a star-connected
slip-ring rotor with a transformation ratio of 3.6 (stator/rotor). The rotor resistance is
0.1 /phase and its per phase leakage inductance is 3.61 mH. The stator impedance
may be neglected. Find (i) the starting current and torque on rated voltage with short-
circuited slip-rings, and (ii) the necessary external resistance to reduce the rated
voltage starting current to 30A and the corresponding starting torque.
Q.49 An ac operated universal motor has a 2-pole armature with 960 conductors. At a
certain load the motor speed is 5000 rpm and the armature current is 4.6A; the
armature terminal voltage and input are respectively 100 V and 300 W. Calculate the
following quantities assuming an armature resistance of 3.5 .
(i) Effective armature reactance
(ii) Max. value of useful flux/pole. (8)
Ans:
The operating conditions in terms of voltage and current of the armature circuit are
shown in the Fig :
76
Ea is in phase with Ia
Or 65.2 + j75.8 - Ea = 16.1 + j 4.6 Xa
Q.50 Define voltage regulation of a single phase transformer. The primary and secondary
winding of a 40kVA, 6600/250V, single phase transformer have resistance of 10 ohm
and 0.02 ohm respectively. The total leakage reactance is 35 ohm as referred to the
primary winding. Find full load regulation at a pf of 0.8 lagging. (8)
77
Q.51 A star connected synchronous motor at 187 kVA, 3-φ, 2300V, 47A, 50Hz, 187.5 rpm
has an effective resistance of 1.5 ohm and a synchronous reactance of 20 ohm per
phase. Determine internal power developed by the motor when it is operating at rated
current and 0.8 power factor leading. (6)
78
Let the additional resistance required in armature current be of R ohms to reduce the
speed to 800 rpm when the load torque is proportional to speed.
T2 = T1 N2/ N1 = T1 800/ 1000 = 0.87
Or Ia2 Φ2 = 0.8Ia1 Φ 1
Or Ia3 Φ3 = 0.64Ia1 Φ1
79
Drop in line V = VS - VR
VS - VR = RP + XQ / VR
80
81
DESCRIPTIVES
Q.1 Draw and explain the phasor diagram of a transformer on load at a lagging power
factor.
(7)
Ans: (D)
Fig. AA1
Fig. AA2
Fig AA2 shows the phase diagram of a transformer on a load at lagging power factor
and corresponds to the equivalents circuit of transformer (Fig. AA1) in which all
quantities are referred to the primary. Thus V2 is the secondary terminal voltage
referred to the primary [where over bar implies a phasor].
2
N1
R = Req = R1 + R2′ = R1 + R2
N2
Where R1 and R2 are primary and secondary resistances, N1 and N2 are primary and
secondary number of turns and X l1 and X l 2 are leakage reactances of primary and
secondary windings.
82
Also
2
N
X = X eq = X l1 + X l′2 = X l1 + 1 X l2
N2
The voltage V1 which is applied to the primary can be written as
V 1 = V 2 + I (R + jX )
Q.2 Explain with proper phasor diagrams the operation of a 3 phase synchronous
machine with normal excitation at the following conditions:
(i) The machine is floating on the supply bus. (7)
(ii) The machine is working as a synchronous motor at no load. (7)
Ans:
83
Fig B1 shows the circuit diagrams and phasor diagrams of a synchronous machine at
generating mode [Fig B1(a)& (c)] and motoring mode [Fig B1(b)&(d)] . The machine
is assumed to be connected to an infinite busbar of voltage Vt and its resistance is
taken to be zero. It is seen from [Fig B1(c)] that in the generating mode, the excitation
emf Ef leads Vt by an angle δ ; on the other hand in the motoring mode Ef lags Vt.
It can be inferred from the above that, when the machine is floating, it is working
neither as a generator nor as a motor and Ia will be equal to zero. E f and Vt will be
equal and have the same phase [Fig. B2]
When the machine is working as a synchronous motor at no load, the figures B1 (b)
and (d) will hold good, but with a very small value for the current Ia; correspondingly
the angles δ and φ will be small. The reason for this is, that at no load the machine
will take that much power which is the sum of friction and windage losses. [Fig B3]
and is small. jI a X s will also be small.
84
Q.3 Draw the torque speed characteristics of a 3 phase induction motor and clearly
indicate the effect of change in rotor resistance. (7)
Ans: The Thevenin equivalent of an induction motor circuit model is given in Figure
D1
R 2′
3V T h 2
s
T = 2
__________
__________
____
( A)
R′
+ ( X T H + X 2′ )
2
ωs RT H + 2
s
85
(1) The starting torque is affected: it increases with increase in R2, goes to a
maximum of Tmax and then decreases with further increase in R2.
(2) Tmax, remains the same but the slip at which it occurs is changed with R2.
Q.4 For a small and sensitive servo mechanism give four reasons why a.c. servo motors
are generally preferred to d.c. servo motor. (7)
86
Fig. E3
The two phase ac servomotor [Fig E2] on the other hand is ideally suited for low
power control applications. The two phases are called control phase (phase a) and
reference phase (phase m), the latter being excited at a fixed magnitude of
synchronous a.c. voltage, both voltages being taken from the same source. The control
phase voltage is shifted in phase by 90 from the reference phase voltage by means of
phase-shifting networks. The motor torque gets reversed by phase reversal of the
control phase voltage.
(iv) it has no brushes that contact the commutator segments. The rotor can withstand a
higher temperature as it does not involve any insulation.
87
Ans: In resistance welding a heavy current is passed through the joint to be welded
and the heat caused by the resistance of the joint is sufficient to cause fusion of the
metal. Three types of resistance welding exist and they are as follows:
(i) Butt Welding:This method is used for welding of rods, wires or small pipes;
the two ends are pressed together mechanically to form a butt joint as shown
in Fig G1
(ii) Spot Welding: For jointing two or even three sheets of metal by means of an
overlapping joint, as shown in Fig G2, The sheets are held between two
electrodes and current is passed between these electrodes, causing fusion at a
single spot.
Q.6 What are different welding controls used in resistance welding? (4)
Ans: The three different controls commonly in use are as follows:
(i) Mechanical Control for Constant – time equipment: A cam- operated
switch connected in the primary circuit of the welding transformer and driven
from the welding machine provides a simple device.
88
(iii) Current and energy- actuated control: With this control, current is allowed
to flow until a predetermined amount of energy has been supplied to the weld.
Q.7 Explain the principle of high frequency induction heating. What factors control the
depth of penetration of heat? Give the industrial application of this mode of heating.
(7)
Ans: Induction heating processes make use of currents induced by electromagnetic
action in the material to be heated. Sufficient currents that cause effective heating can
be produced only in materials of low resistivity, however it is necessary to use a
magnetic field of very high frequency.
For coreless induction furnaces, the depth of penetration(t) is given by the formula
1 ρ .109
t=
2π µf
f =frequency(Hz)
and µ =permeability (this factor has a value of 1.0 for molten steel)
For normal supply frequency, with the power factor in the range 0.8-0.85 and in sizes
upto about 1 tonne, the vertical core-type furnace is widely used in foundries for
melting and refining brass and other non –ferrous metals. On the other hand the
coreless induction furnace is used for the production of very high grade alloy steels; in
small sizes it is widely used for work on alloys and precious metals.
Q.8 What is the fundamental difference between thermal and nuclear power plants? (3)
Ans: In a thermal power plant heat is released in combustion of coal; this heat is used
in a boiler to raise steam.
Here the coal is conveyed to a mill and crushed into a fine powder, this being termed
as pulverization. The pulverized coal is blown into the boiler where it mixes with the
supply of air for combustion. Heat is transferred to steam pipes located in the top
region of the boiler, these being initially fed with hot water from the boiler feed
pump; the hot water then gets converted to steam at high temperature and pressure.
This steam is fed to the steam turbines which are the prime movers for electric
generators.
89
Q.9 List the advantages of nuclear power plants over conventional thermal power
plants. (4)
Ans:
(ii) It requires very little fuel in terms of volume and weight and therefore poses
no transportation problems. The nuclear power plant may be sited,
independently of nuclear fuel supplies, close to load centres. However safety
considerations require that these plants be normally located away from
populated areas.
(iii) When one pound of pure uranium is completely fissioned it will create as
much heat as the burning of 1500 tons of coal.
(iv) Whereas the stock of coal is limited and the supply of coal will go down in the
coming decades, it is the nuclear energy that is promising. Nuclear power
plants will therefore be important sources of electrical power in future.
Q.10 Discuss briefly the solid state circuits used for the stator voltage control of
induction motor derives. (7)
Ans : Fig H1 shows the block diagram for a scheme of stator voltage control. Here
the ratio V/f is kept constants, with the frequency being varied by means of an
inverter. This in turn will help in maintaining the field flux constant. The rectifier and
inverter are rigged by means of the thyristors or GTOs or MOSFETs. Because of its
high cost, this type of control is justified only for drives wherein rugged,
maintenance-free characteristics of the induction motor are essential.
90
91
Q.12 What are the advantages of high voltage transmission? Give its limitations also.
(4)
Ans: High voltage transmission is subdivided into HVAC and HVDC transmission
systems.
(i) HVAC transmission: Advantages of HVAC transmission are as follows:
As the voltage is increased, the current carried by the conductors decreases. The i2R
losses correspondingly get reduced. However the cost of transmission towers,
transformers, switches and circuit breakers rapidly increases with increase in voltage,
in the upper ranges of a.c. transmission voltages.
(ii) HVDC transmission:
Advantages They (HVDC lines) are economical for bulk power transmission. The
voltage regulation problem is much less in DC since only IR drop is involved. There
92
is easy reversibality and controllability of power flow through a DC link. Also there is
considerable insulation economy.
Limitations: The systems are costly since installation of complicated converters and
DC switchgear is expensive. The converters require considerable reactive power.
Lack of HVDC circuit breakers hampers network operation. Moreover there is
nothing like DC transformer; voltage transformation has to be provided on the a.c.
sides of the system.
Ans : Overcurrent relay: This relay operates when current through it (sample of
power system current) satisfies the condition
2
Q = K1 I − K 4 > 0
K4
or I > = Ip
K1
where I p is said to be the pickup value of the relay. Such a relay is an over current
relay and would operate in the shortest possible time (depending upon the type of
hardware employed), and is called instantaneous overcurrent relay.
Thus if:
I > I p the relay trips the circuit broker
and
if I < I p the relay blocks or does not trip the circuit breaker.
The directional feature can be incorporated for the above o/c relay by means of an
induction cup type of structure (Fig L1) one of the coils being current excited and the
other, voltage excited.
93
The phasor diagram of a directional relay is shown in Fig. L2. I v , the voltage coil
current lags the voltage applied by an angle depending upon the coil impedance. Since
the fluxes are proportional to the coil currents, and have the same angle separation as
coil currents, the relay torque can be expressed as
94
T = K 3 V I sin(θ − φ ) (A)
Taking the leading angle as positive and defining τ (a relay parameter) as the value
of θ , when the relay develops maximum torque, then
τ −φ = 90
Substitution of this relation in Equn (A) gives
T = K 3 V I cos(θ − τ )
which indeed is the directional characteristic
Q.14 Explain the working of nickel cadmium cells with its merits and demerits over lead
acid cell. (7)
Ans: Nickel- Cadmium cells: In a Nickel-Cadmium cell the positive plates are made
of nickel hydroxide enclosed in finely perforated steel tubes or pockets, the electrical
resistance being reduced by the addition of flakes of pure nickel or graphite. These
tubes or pockets are assembled in nickelled-steel plates. The active material in this
cell is Cadmium mixed with a little iron, the latter being used to prevent caking of the
active material and hence an impairment to porosity.
Chemical reaction
Positive Negative Positive Negative
plate plate plate plate
95
96
97
In the single phase reluctance motor the rotating field can be produced by one of the
phase-splitting methods. The salient pole structure is given to the rotor by removing
some of the teeth of an induction motor rotor as shown in Fig M2. The remaining
teeth carry short-circuited copper bars to provide the starting induction torque. After
starting, the rotor reaches near synchronous speed by induction action and is pulled
into synchronism during the positive half-cycle of the sinusoidally varying
synchronous torque.
98
This would only be possible if the rotor has low inertia and the load conditions are
light. The torque speed characteristic of a typical reluctance motor with induction start
is given in Fig M3. Here the starting torque is highly dependent upon the rotor
position because of the projecting nature of the rotor. This phenomenon is known as
“cogging”. For satisfactory synchronous motor performance the frame size to be used
must be much larger than that for normal single- phase induction motor. This accounts
for the high value of starting torque shown in Fig M3.
(ii) Parallel operation of transformers: when the load exceeds the capacity of an
existing transformer, it may be economical to install another one in parallel rather
than replacing it with a single larger unit. Also for reliability two smaller units in
parallel are preferred. The cost of maintaining a spare is also less with two units in
parallel.
For satisfactory parallel operation of transformers the following conditions have to be
fulfilled:
(i) They must be connected with proper polarities; this is to ensure that the net
voltage around the local loop is zero. A wrong polarity connection results in a dead
short circuit.
(ii) 3 phase transformers must have zero relative phase displacement on the secondary
sides and must be connected with proper phase sequence. Only the transformers of the
same group can be paralleled. For example Y/Y and Y/ ∆ transformers cannot be
paralleled as their secondary voltages will have a phase difference of 300
(iii)The transformers must have the same voltage ratio to avoid no-load circulating
current, when the transformers are in parallel on both primary and secondary sides.
(iv) There should exist only a limited disparity in the per unit impedances (on their
own bases) of the transformers. The currents carried by the two transformers are
proportional to their ratings if their p.u. impedances on their own ratings are equal.
Fig N1 shows two transformers paralleled on both sides with proper polarities but on
no load. The primary voltages V1 and V2 are equal. If the voltage ratios of the two
99
transformers are not identical, the secondary voltages E1 and E2 though in phase will
not be equal in magnitude. The difference (E1-E2) will appear across the switch S.
When the secondaries are paralleled by closing the switch, a circulating current
appears even though the secondries are not supplying any load. The circulating
current will depend upon the total leakage impedance of the two transformers and also
the difference in their voltage ratios. Only a small difference in the voltage ratios will
be tolerated.
Division of load between transformers in parallel. Equal voltage ratios When the
transformers have equal ratios E1=E2 in Fig N1, the equivalent current of the two
transformers would then be as shown in Fig N2 on the assumption that the exciting
current can be neglected in comparison to the load current.
100
also I 1 + I 2 = I L
Z2
S1 = SL
Z1 + Z 2
Z1
S2 = SL
Z1 + Z 2
where S1 = VL * I1
S2 =VL *I 2
SL = VL * IL
The above equations S1 , S2 and S L are phasor relationships giving loadings in the
magnitude and phase angle.
(iii) Selsyns: Selsyns or synchros are control system components which are used for
transmission of small torques or motions electrically. They can be categorize into
three kinds:
(a) for transmission of small torques electrically, as sinchro-transmitter-receiver
pair (Fig P2)
(b) for indicating the difference in positions, as generator – transformer
combination (Fig P3)
(c) as differential selsyns(Fig P4)
In the synchro-transmitter-receiver pair (Fig P2) which is a single phase
selsyn, the stator has three windings like the polyphase induction motor. The
rotor is similar to the rotor of a small alternator and of one winding. Fig P1
shows the cross –section of a single phase selsyn.
101
Although it shows three stator windings, it is still a single –phase device. Any
a.c. current in the rotor will produce at “stand–still” three stator voltages
which are in time phase. Two of these devices in the circuit of Fig P2 provides
a system for transmission of motion.
It is assumed that the generator and motor are similar units and that for the initial
conditions the voltages produced on the stator of the generator by the generator rotor
are equal in magnitude and 1800 out of phase with those produced by the motor rotor
in the motor stator. Under these conditions the stator currents will be zero and no
torque will be present in either machine.
If the rotor of the generator is turned through an angle θ while the position of the
rotor of the motor is left unchanged, a circulating current Ia will result in the stators.
The current acting on the air gap flux will tend to restore the generator to its original
position. The current will produce a torque in the motor which wil tend to cause the
rotor of the motor to assume an angle corresponding to that occupied by the rotor of
102
the generator. If the rotor of the motor is free to turn, it will follow the angular
position of the generator rotor. The rotor of the motor therefore is an indicator of the
position of the generator rotor or any device connected mechanically to it.
Selsyns as position indicators: If the circuit to the rotor of the motor is opened (Fig
P3) a voltage will be produced in the rotor winding, the magnitude of which is a
function of the angular position of the rotor of the motor w.r.t. that of the generator
rotor. When the motor rotor is at a position of 90 electrical degrees from the position
occupied and the rotor is electrically connected and free, the magnitude of the rotor
voltage will be zero. Then any movement of the generator rotor will produce a voltage
in the motor rotor which is a function of the angular position of the generator rotor (or
other equipment coupled to it). This circuit has found use in many servomechanism
systems. For this application the motor unit serves just as a control transformer.
A third useful application of the selsyn system is that of the differential selsyn shown
in Fig P4; this is constructed like a wound rotor induction motor having a 1:1 ratio of
turns. When the rotors of the motor and generator of Fig P3 are in a given position,
the differential selsyn will adjust the position of the rotor to its stator such that
minimum current flows in its windings. With the rotor of the motor locked in a
particular position, a change in the position of the generator rotor will cause a
corresponding change in the position of the rotor of the differential selsyn. In this way
the circuit of Fig P4 performs the same function as that shown in Fig P3.
103
Q.17 Give the layout of a hydro – electric power plant and briefly explain the working of
any two of its main components. (14)
Ans: A typical hydroelectric development consists of a dam, which raises the water
surface of the stream to create head, pondage, storage or the facility or diversion into
conduits, an intake containing racks and gates to control the flow into the conduits, a
conduits system to convey the water to the turbines which transform the energy of the
water into mechanical energy, generators to transform the mechanical energy of the
hydraulic turbines into electrical energy, a power house to contain the turbines,
generators and accessories and a draft tube to convey the water from the turbine to the
tailrace which leads back to the stream.
104
From the reservoir, the water flows through the intake 1 and penstock 2 to the
hydraulic turbines 3. On flowing through the turbines where it gives up its energy ,
the water flows through the draft tube to the tail race. The turbine rotates the rotor of
the generator 4. As a result, electric current is induced in the generator stator 5. The
current is conveyed via the buses 6 to the step-up transformer 7, and then to the
switchgear 8. From the switchgear, the current is directed through the power lines 9 to
the loads.
Two of its main components are described below:
Hydraulic Turbines
A hydraulic turbine is a prime mover which transforms the energy of falling water
into mechanical energy of rotation and whose primary function is to drive a water
wheel hydroelectric generator. The turbine runner and rotor of the water wheel
generator are usually mounted on the same shaft, which is why the entire assembly is
frequently referred to as the turbo- generator.
An impulse turbine is one in which the driving energy is supplied by the water in
kinetic form; and a reaction turbine is one in which the driving energy is provided by
the water partly in kinetic and partly in pressure from.
Switchgear
The functions of the switchgear may be briefly summarized as follows
(i) To localize the effects of faults by operation of protective equipment and to
automatically disconnect faulty plant from the system.
(ii) To break efficiently, short-circuits without giving rise to dangerous conditions.
(iii) To facilitate re-distribution of loads, inspection and maintenance of the
system.
The design and construction of the switchgear layout should be such that a
reliable service is obtained under all operating conditions and in, as direct a
route as possible from the alternators to the outgoing feeders. The choice of
suitable switchgear is governed by the maximum short circuit-MVA it is
called upon to deal with and also to some degree upon its relation to the
system of which it forms a part. The figure of short circuit MVA is needed for
installing suitable circuit breakers.
105
2
K1 I 2 + K 2 V > 0
or
1 2
V K1
= Z < = Z rs
I −K2
106
If K2 is made negative, such a relay senses impedance magnitude, and operates if the
magnitude of impedance seen from its location(in any direction) is less than a
specified value. Since V prevents relay operations, K 2 being negative, and I tries
to operate it, voltage is the restraining quantity and current is its operating quantity.
A modified impedance relay called mho relay results if a directional relay is
restrained by voltage. Thus
2
Q = K2 V I cos(θ −τ ) − (−K1) V − K4
or
V K3 ___________
(1)
= Z < cos(θ − τ )
I − K1
The right hand side of Eqn (1) is a circle with centre located in the line determined by
the parameter τ and passing through the origin as shown in Fig D3. The characteristic
in this case is inherently directional. This characteristic can be alternatively expressed
as
Z − Z r s < Z r s : tr ip
Z − Z rs > Z rs : block
107
(iv) What is welding process and explain principle of electric welding. (14)
Ans:
(i) Servomotor: Servomotors are of two kinds, d.c. and a.c. These are described
below:
The two-phase ac servomotor [Fig E4] on the other hand is ideally suited for
low power control applications. The two phases are called control phase
(phase a) and reference phase (phase m), the latter being excited at a fixed
magnitude of synchronous a.c. voltage, both voltages being taken from the
same source
108
The control phase voltage is shifted in phase by 90 from the reference phase voltage
by means of phase shifting networks. The motor torque gets reversed by phase
reversal of control phase voltage. This motor has low rotor inertia and hence faster
response; also it has no brushes that contact the commutator segments.
109
• Surface hardening
• Deep hardening
• Tempering
• Soldering
• Melting
• Smelting etc.
(iv) Electrical welding
Welding is the process of joining metals of similar composition by heating to suitable
temperature with or without application of pressure and addition of filter material.
The result of welding is a continuity of the homogeneous material and is of the same
composition and characteristics as the parts, which are joined together.
Welding process may be classified as :
(i) Pressure welding: The process in which the metal parts to be joined are heated
to a plastic stage and then joined by applying external forces are known as plastic or
pressure welding.
(ii) Fusion welding: The process in which the metal parts to be joined are heated
to molten state and then allowed to solidify joining a localized homogeneous union of
the two is known as fusion welding.
Welding process may be further classified as:
a. Electric resistance welding: Electric resistance welding is most
commonly used kind of pressure welding. It can be applied to almost any metals with
in certain limits of composition.
110
111
We also know that when the current passes through any conductor, there is loss of
power in that particular conductor according to the relation, I2 R. As the loss is
proportional to the square of the current. So if the current is reduced to ½ value , then
the loss will be reduced to 1/4th its original value. Hence the efficiency of the
transmission line and all others equipments associated with the line will increase and
more power will be available for use.
When current is passing through a conductor there will be a voltage drop according to
the relation V=IR. So, when the current is reduced the drop of the voltage is less in
the line, of course with the same cross sectional area of the conductor.
With the reduction of cross sectional area, considered the main advantage of
transmitting electrical energy at very high voltage viz 132kV , 220kV or even 400kV.
But in case of distribution system such high voltage is dangerous, so distribution
voltage is generally 400/230V.
112
signals and the modulated signal is fed to the amplifier and is then transmitted via
coupling unit.
Q.21 What are the various types of electric drives? Compare their advantages and
disadvantages. State and explain the various factors which affect the selection of an
electric motor as industrial drive. (14)
Closed-loop control of d.c. drives usually involves one or more feddback loops. The
relevant signals are sampled through transducers at discrete instants for computational
purposes; also switching of power semiconductor devices is accomplished by
gate/base control. Fig M14 shows a block–cum schematic diagram of a typical
separately excited d.c. drive which has an outer speed loop and an inner current
control loop. The motor speed is sensed by a tachometer, the output of which is fed
through an A/D converter to the microcomputer. However a shaft encoder can also be
used for this purpose, and the digital output of the shaft encoder can be directly fed to
the microcomputer. The armature current of the motor can be sensed by measuring the
voltage across a standard resistor.
113
dc Field
motor
A/D Field
power
converter converter
Microcomputer
Current
sensing
Technogenerator Single-
phase
ac supply
Fig. M14 Block diagram of a separately excited dc drive
Till recent times variable speed drive application were dominated by d.c.
motors. Squirrel cage induction motor which costs approximately one – third
of a d.c.-motor of the same rating, is quite rugged, maintenance free, can be
built for higher speeds, torques and power ratings. Wound rotor induction
motors, through costlier than squirrel cage ones, also need less maintenance
and are available in higher power ratings. Thus a.c. drives have succeeded in
replacing dc drives in a number of variable speed applications. For control of
a.c. motors that include both types of induction motors as well as synchronous
motors fed from d.c. supply, thyristor, transistor or MOSFET based inverters
are employed. These semiconductor switching circuits transfer energy from dc
supply to an a.c. load of variable frequency and/or variable voltage. The
switching operation results in considerable harmonics in the input a.c.
waveforms which are filtered out by the a.c. motor. The normal power supply
being of the a.c. type, the complete scheme involves a converter and then an
inverter.
A cycloconverter is a direct a.c. to a.c. converter for obtaining variable
voltage low frequency a.c. supply from a fixed frequency one. Thus it
eliminates the necessity of on intermediate d.c. stage. Inspite of their direct a.c
to a.c. conversion the cycloconverters have the disadvantages that
(i) They produce a subfrequency ouptput,
(ii) their output contains a high harmonic content, and
114
(iii) the input power factor is low. Cycloconverters are used for low-speed
drives and controlling linear motors in high-speed transportation systems.
Ans:
(i) Induction Furnace: Induction heating processes make use of currents induced
by electro-magnetic action in the material to be heated. It is, however only in
materials of low resistivity such as metals that sufficient currents to cause effective
heating can be produced. For this purpose it is often necessary to use a magnetic field
of very high frequency. For melting or refining metals, various kinds of induction
furnaces are available, while for other purposes such as case hardening and soldering,
the necessary currents are induced in the article at the appropriate stage of its
manufacture; the latter process requires a high frequency and is called as the high-
frequency eddy current heating.
The core type furnace is essentially a transformer in which the charge to be heated
forms the secondary circuit and is magnetically coupled to the primary by an iron-
core as shown in Fig N1. It can be seen from this figure that the magnetic coupling
between the primary and the secondary is poor. The electromagnetic forces cause
turbulence of the molten metal which, although useful upto a point, is liable to
become to severe. A crucible of inconvenient shape from metallurgical point of view
is used. In order to start the furnace, a complete ring of metal must be present in the
crucible, otherwise the secondary circuit will not be complete.
115
Chore
Cone
An improvement over the conventional core type furnace is the vertical core type
furnace which overcomes some of the above mentioned difficulties. It is known as the
Ajax-Wyatt furnace [Fig N2] and uses a vertical instead of horizontal channel for the
charge. The tendency of the circuit to repture due to the pinch effect is counteracted
by the weight of the charge in the main body of the crucible. The circulation of
molten metal is kept up round the Vee portion by convection currents and by the
electromagnetic forces between the currents in the two halves of the Vee. Such a
furnace is very widely used in foundries for melting and refining brass and other not-
ferrous metals.
(ii) Buchholz’s relay : The Buchhloz’s relay is a gas operated relay in which the
development of a fault within the transformer, particularly in the incipient stage is
guarded. Also major breakdowns and sudden disruption of supply are avoided, thus
making it a very reliable piece of electrical equipment. It is fitted to transformers
having conservator vessels and is installed in the pipeline between the transformer and
its conservator tank. The relay comprises an oil tight container fitted with two internal
floats which contain mercury switches connected to external alarm and tripping
circuits. Normally the device is full of oil and the floats, due to their buoyancy rotate
on their supports until they engage their respective stops.
116
a) An incipient fault within the transformer generates small bubbles of gas which
pass upward, towards the conservator, become trapped in the top of the chamber,
thereby, causing the oil level to fall. The upper float gets deflected as the oil level
within the relay falls and when sufficient oil has been displaced, the mercury
switch contacts close, thus completing the electrical alarm circuit.
GAS AND OR
ACTUATED RELAY
b) For a serious fault within the transformer, the gas generation is more violent and
the oil displaced by the gas bubbles flows through the connecting pipe to the
conservator. This abnormal flow of oil causes the lower floats to be deflected thus
actuating the contacts of the second mercury switch and completing the tripping
circuit of the transformer circuit breaker, thus disconnecting the transformer from
the supply. The arrangement for mounting the gas and oil actuated relay is shown
in Fig N3.
117
the conductors of a three phase line with unsymmetrical spacing. For the purpose of
this discussion it is assumed that there is no neutral wire so that the conductor currents
sum upto zero. That is
This arrangement causes each conductor to have the same average inductance over the
transposition cycle. It can be shown that over the length of one transposition cycle, the
total flux linkages and hence the net voltage induced in a nearby telephone line is zero
and also that the power line transposition is ineffective in reducing the induced
telephone line voltage
i) when the power line currents are unbalanced (that is the sum I a + I b + I c ≠ 0 ) or
ii) when they contain third harmonics. These harmonic line currents are undesirable
because:
a) the induced emf becomes proportional to the frequency and
b) higher frequencies come within the audible range. Thus there a need to avoid the
presence of such harmonic currents on a power line from considerations of the
118
From figure, flux increases from its zero value to maximum value Φm in one quarter of the
cycle i.e. 1/4f second
Average rate of change of flux = Φ m /(1/4f)
= 4 f Φ m Wb/s
Rate of change of flux per turn means induced emf in volts
Average emf/turn =4 f Φm
If flux Φ varies sinusoidally, then rms value of induced emf is obtained by multiplying the
average value with form factor.
119
Q.24 Draw the characteristic curves and state two applications for
(i) a dc shunt motor
(ii) a dc series motor. (4)
Ans:
3(a) Characteristic curves & applications
D.C. shunt motor :-
120
Applications :- Due to the constancy of their speed shunt motors are suitable for driving
shafting, machine tools, lathes, wood working machines and for all purposes where an
approximately constant speed is required.
DC Series Motor:-
Applications:- Traction work, trolley cars, hoists & cranes, conveyors — etc.
121
Q.25 With a neat diagram explain the working of a universal motor. Also draw its
torque-speed characteristics when it is fed from both ac & dc sources. (7)
Ans:
4. (a) Universal motor :-
Q.26 Explain the construction, working and applications of a stepper motor. (7)
Ans:
Stepper motor :-
Construction: Stepper motor consists of a slotted stator equipped with two or more
individual coils and a rotor structure that carries no winding .The classification of the stepper
motor is determined by how the rotor is designed. If the rotor is made up of a permanent
magnet, it is called a PM stepper motor. If no magnet exists on the rotor (only a rotor core) it
is called a reluctance type stepper motor. The presence of the permanent magnet in the rotor
furnishes the motor with the equivalent of a constant DC excitation.
122
Operation:-
The elementary operation of a four phase stepping motor with a 2-pole rotor is explained
below.
With the above construction of a stepper motor the rotor assumes the angles θ = 0°, 45°,
90°, 135°, 180°…etc. as the windings are excited in the sequence of Na, Na+ Nb, Nb, Nb+ Nc,
Nc…etc.
The same motor can also be used for 90° steps by exciting the coils singly i.e. Na, Nb,
Nc , Nd only. For the use of 90° steps only the permanent magnet rotor can be used.
Applications :-
(1) Typewriters, printers
(2) Positioning of disk drives
(3) Plotters
(4) Recording heads in computer disk drives
(5) Worktable and tool positioning in CNC machines.
Major advantage of using stepper motors is that no feedback is required from the rotor
to ascertain the angular position of rotor.
Q.27 Compare the induction motor with a transformer. (2)
Ans:
5(a) Comparison of Induction Motor with a transformer :
123
The transfer of energy from stator to the rotor of an induction motor takes place entirely
inductively with the help of flux mutually linking the two. Hence an induction motor is
essentially a transformer with stator forming the primary and rotor forming the rotating
secondary.
Transformer Induction Motor
(1) Secondary is stationary (1) Secondary winding is rotating
(2) Secondary is not short circuited (2) Secondary is always short circuited
(3) No-load current is about 1% of full load (3) No-load current is approximately 30 to 50%
current (due to low reluctance path of full current (due to high reluctance
of steel core) of air gap)
(4) emf induced in secondary depends on (4) Depends on K and slip also
K (turns ratio)
(5) Frequency of primary and secondary currents (5) Frequency of stator current (f)and rotor
are same current (sf) are not the same.
(iii) A C servomotors.
OR
Switched reluctance motors. (5+5+4)
Ans:
For power transformers frequency response is of no serious interest because these units are
operated at a single fixed frequency. Frequency response is to be studied in communication
circuits where the frequency of source voltage is likely to vary.
124
Equivalent circuit :
Frequency Response :
Fig (a)
125
Fig (a)
Fig (b)
Fig (c)
126
When only IF is present the, mmf is perpendicular to MNA (Figure (a). When only
armature current is present, mmf is along the MNA. When the machine is actually working,
the resultant mmf vector is at an angle form MNA. This causes the brushes to be shifted to a
new MNA perpendicular to resultant mmf. Now at this new MNA, mmf of Ia is along new
MNA. This mmf is resolved into two component Fd and Fc which are demagnetizing and
cross magnetizing mmf respectively.
127
(iii) AC Servomotors :
Generally a large X/R ratio is used in induction motor to have high torque at the
operating region i.e. around 5% slip. But in a servomotor, for a small X/R the characteristic is
nearly a linear in contrast to high nonlinear characteristic of large X/R. When large X/R is
used for servo application because of the positive slope for part of characteristic (a) the
system using such a motor becomes unstable.
OR
128
into the densest portion of the field. The force tends to align the specimen of material
in such a way that the reluctance of the magnetic path that lies trough the material will be
minimum.
When the stator winding is energized the revolving magnetic field exerts reluctance
torque on the unsymmetrical rotor tending to align the salient pole axis of the rotor with axis
of the RMF (due to minimum reluctance). If the reluctance torque is sufficient to start the
motor and its load, the rotor will pull into step with the rotating field and to run at the speed
of the revolving field.
The switched reluctance machine is a special variation of the simple reluctance machine
that relies on continuous switching of currents in the stator to guarantee motion of the rotor. It
is also a true reluctance machine in that it has salient poles both in rotor and in the stator.
Q.29 What do you understand by electric heating? State its advantages over other
methods of heating. (7)
Ans:
Electric heating
Electric heat is provided by use of a high temperature wire, which is resistant to electric
flow with more heat being produced by more electric current being fed into the heating
element, creating more resistance and more heat. This method of producing heat is called
electric heating.
1. Cleanliness
5. Quicker operation
6. Higher efficiency
129
Q.30 Describe the resistance welding method and name the other welding methods. (7)
Ans:
Types of Welding :
Resistance Welding :
Resistance welding is a process used to join metallic parts with electric current. In all forms
of resistance welding, the parts are locally heated until a molten pool forms. These parts are
then allowed to cool and the pool freezes to form a weld nugget. To create heat electric
current is passed through work pieces. The heat generated depends on the resistance and
thermal conductivity of the metal and the time for which the electric current is applied. The
heat is expressed by the equation.
E= I2Rt
E is the heat energy
I is the current
R is the electrical resistance
t is the time for which the current is applied.
Q.31 List out the advantages of nuclear power plants over conventional thermal power
plants. (4)
Ans:
Advantage of Nuclear Power Plants
1. Nuclear Power station reduces the demand for oil coal and gases.
2. Weight of nuclear fuel required for a given station capacity is negligible compared to
that required for a conventional thermal power station. As a result the problems of
transport and resulting costs are avoided.
3. The area and volume need for a nuclear power station is less in comparison to a
conventional power plant of equal capacity.
4. In addition to producing large amount of power, a nuclear power plant produces
valuable fissile material, which is extracted when the fuel is renewed.
130
Q.32 Draw a flow diagram of a thermal power station and explain its main components.
(10)
Ans:
Flow diagram of thermal power station :
131
Q.33 Draw the per phase equivalent circuit of a 3 phase induction motor and derive the
expression for torque. (2)
Ans:
(a) Per phase Equivalent circuit of Induction Motor
Torque Equation:
Rotor input = torque x ws
Torque/phase = rotor input / ws = rotor cu loss/ (ws s)
132
Q.34 What do you mean by transmission line efficiency and regulation? Can it be
negative also? (4)
Ans:
Transmission Efficiency:-
The ratio of receiving end power to the sending end power of a transmission line is
known as the transmission efficiency of the line.
% Transmission Efficiency =(Receiving end power /Sending end power) x 100;
= (VR IR cosφR/ VS IS cosφS) x100;
where VR ,IR and cosφR are the receiving end voltage, current and power factor while VS, IS
cosφS are the corresponding values at sending end.
Voltage Regulation :-
The difference in the voltage at the receiving end of a transmission line between the
conditions of no load and full load is called voltage regulation and is expressed as a
percentage of the receiving end voltage.
Mathematically,
% voltage regulation =(VS –VR)/VR x 100;
%Voltage regulation is also given by= (I R cos φR ± I X sin φR)/ER x100;
+ for lagging pf
– for leading pf;
When the pf is leading, and the term IX sin φR is more than the I R cosφR, then the voltage
regulation becomes –ve.
Q.35 With the help of neat sketches explain different types of distribution systems. (10)
Ans:
Different types of distribution systems :-
(i) Radial system :
In this system separate feeders radiate from a single sub station and feed the distributors at
one end only. Figure shows a single line diagram of a radial system for a DC distribution
where a feeder OC supplies a distributor AB at a point A. Obviously the distributor is fed at
one end only i.e. point A in this case.
The radial system is employed only when power generated at low voltage and sub-station is
located at the center of the load.
133
The sub-station supplies to the closed feeder LMOPQRS. The distributors are tapped from
different points M, O, and Q of the feeder through distribution transformer. This system has
the advantage of continuity of supply and less voltage fluctuations.
When the feeder ring is energized by two or more than two generating stations or substations
it is called inter connected system.
134
In the above diagram of the inter-connected system the closed feeder ring ABCD is
supplied by two sub stations S1 and S2 at D and C. Inter connected system has the advantages
of increased service reliability and increasing efficiency of the system by reducing reserve
power capacity.
Ans:
(i) Industrial drive and their control :-
Any motor with a speed control arrangement can be termed as a drive. The application
of these drives for industrial purposes are almost universal since they posses inherent
advantages over other forms of conventional drives viz, it’s cleaner, more easily controllable
and more flexible. Both AC and DC are used for electric drives.
There are two types of drives viz, group drive and individual drive. Each machine
having its own driving motor is called individual drive. When a no. of machines are driven
through belts from a common shaft ,it is called group drive. Selection of the type drive
depends on the application to be served.
Control :- Mostly DC shunt /series motor and induction motor are used as drive motors in
industry.
(1) Control of DC motors
(i) Field control
(ii) Armature control
(iii) Ward-Leonard system of speed control
(iv) Electronic control such as Thyristor control, Chopper control
135
(b) Draw in system : In this method, conduit or duct of glazed stone or cast iron or
concrete are laid in the ground with manholes at suitable positions. The cables are then
pulled into position from manholes. Figure shows section through four way underground
duct line. Three of the ducts carry information cables and fourth duct carries relay
protection connection, pilot wires. Care must be taken that where the duct line changes
the direction, depths, dips and offsets are made with a very long radius or it will be
difficult to pull a very large cable between manholes.
136
(c) Solid system : In this method of laying the cable is laid in open pipes or troughs
dug out in earth along the cable route. The troughing is of iron, stone ware, asphalt or
treated wood. After the cable is laid in position, the troughing is filled with a bituminous
or asphaltic compound and covered. Cables laid in this manner are usually plain lead
covered because troughing affords good mechanical protection.
On long lines carrier pilot relaying is cheap and more reliable than the pilot wire
relaying even though it is expensive and complicated than with pilot wire relaying.In this
type of protection the phase angle between two ends of the line decides whether the fault is
internal or external. When a voltage of positive polarity is impressed on the control circuit , it
generates a high frequency o/p between on phase conductor and ground. Each carrier current
receiver receives carrier current from its local transmitter as well as from the transmitter at
the other end of the line. These signal are then compared for any fault condition and action is
taken correspondingly. Traps are used to confine the carrier currents to the protected section
so as to avoid interference from other carrier current channels.
137
Q.37 Can a transformer be used to transform direct voltage and direct current? Justify
your answer. (5)
Ans: The transformer can not be used for dc supply system (dc voltage & current).
According to working principle of the transformer, it works on Faraday laws of
Electromagnetic Induction. Therefore, induced emf e = (-dΦ /dt). The changing rate of
magnetic flux is responsible for e.m.f generation, which opposes the change of magnetic
flux, that is indicated by negative sign. But in case of dc supply system (d.c Voltage &
Current), there is no change of magnetic flux, dΦ = 0, so that induced emf at secondary
side of transformer is zero. Therefore, we can not use transformer for dc supply system.
Q.38 How does a salient-pole rotor differ from a cylindrical rotor in synchronous
machines. Where are the salient-pole type of synchronous machines used?
(3+3)
Ans:
Q.39 Explain the effect of ‘armature reaction’ on the performance of D.C. motor. (8)
Armature reaction means effect on magnetic field distribution, this works in two ways:
138
Fig - (i)
Fig - (ii)
Fig - (iii)
139
The first effect leads to reduce the speed of dc motor and second to produce sparking
at brushes. For armature reaction illustration, two poles have been considered and
initially assumed that there is no current in the armature conductors. The assumptions
are as under:
- That the flux is distributed symmetrically with respect to polar axis, which is
line joining the centres of NS poles.
- The magnetic neutral axis or plane (M.N.A) coincides with the geometrical
neutral axis(G.N.A)
The figure (i) shows flux distribution by poles (N&S) without armature current and
figure (ii) shows field or flux set up by armature conductors. And figure (iii) shows
the resultant flux or (mmf) component ‘OF’ which is found by vectorally combining
OFm and OFA.
- To transfer power from one circuit to another circuit, used in control and
regulation devices.
140
- Signal Pulse Transformer: These are used for delivery of signal at output.
The transformers deliver pulse like signal or series of pulses. The turns ratio of
pulse Transformer can be used to adjust signal amplitude and provide
impedance matching between source and load.
141
This motor is similar to a 3-phase induction motor except its stator is provided
with a single phase winding and a centrifugal switch is used in some types of
motors, in order to cut out a winding used only for starting purpose. This
motor has distributed stator winding and squirrel cage rotor. When stator is fed
from single phase supply, its stator winding produces a flux which is only
alternating. This alternating or pulsating flux acting on a stationary squirrel
cage rotor cannot produce rotation (only a revolving field can). Therefore,
single phase motors are not self-started.
142
Total torque T = Tf + Tb
Tf = forward torque
Tb = backward torque
To make single phase induction motor self starting, it is temporarily converted
into a two phase motor during starting period. For this purpose, the stator of a
single phase motor is provided with an extra winding known as starting
winding. Two windings are spaced at 900 electrically apart and connected in
parallel across single phase supply.
These motors are widely useful in home, offices, factories/workshop & in
business establishment etc.
143
• Smaller in size.
(b) What are the essential differences between H.V. and L.V. switchgears? (7)
Ans: (a) Different Levels Of Voltages For Generation, Transmission And Distribution:
- The long distance high power transmission is by EHVAC lines 220 kV,
400 kV & 760 kV Ac. In special cases, HVDC line is preffered. The rated
voltages of HVDC lines are ±250 kV, ± 400 kV, ± 500 kV & ± 600 kV.
- Utilization is at the low voltages upto 1kV and medium voltage upto 11 kV.
144
145
Q.45 Why are solid-state controllers preferred over conventional controllers? Give
examples of their applications and explain their features. (14)
Ans: Solid State controllers Vs Conventional Controllers:
Primarly motor control circuit of conventional type consists of switches, fuses, relays,
contactors and associating cabling. This type of conventional contactor control system
become bulky if implemented for complicated logic. The trouble shooting also
becomes difficult and time consuming.
On the other side, solid state controller made up by electronic components / devices
are used to implement such schemes in place of conventional contactor control which
are efficient, reliable and need less time for trouble shooting.
A PLC (Programmable Logic Controller) is known as solid state controller. The main
difference between relay logic controller and computer based solid state controller
(PLC) are:
- All input, output variables are in binary form
- In relay logic programming whenever any event sequence is to be changed,
rewiring of panel is necessary which was cumbersome. In computer based
system, it is easy to change program through software.
- Solid state devices can be easily controlled by computer.
Therefore, Programmable Logic Controller (PLCs) are compact electronics devices
that allow any process logic to be implemented by programming of PLC. It avoids lot
of external control wiring. The PLC can be programmed according to user’s need and
the logic can be changed as per requirement as and when it is required.
146
PLC operation: A PLC work is based on continuous scanning program. It has three
steps
- Check input status.
- Execute program.
- Update output status.
147
The rotor is made by a smooth chrome steel cylinder having high retentivity
so that the hysteresis loss is high. It has no winding. The rotor poles are
magnetically locked up with revolving stator poles of opposite polarity.
However rotor poles lag behind the stator poles by an angle . The mechanical
power developed in a hysteresis motor :
Pm = Ph (1-s)/s where Pm = Mechanical power developed
Ph = Hysteresis loss in rotor
Th = Hysteresis torque
and Th = 9.55 Pm / NS
These motors are useful for driving tape- decks, turn table and other precision
audio equipments. The commercial motors have two poles, they run at 3,000
rpm at 50 Hz , single phase supply.
148
149
- Electrolyte : Alkali or organic, alkali required for zinc cathode and organic
for lithum cathode.
- Nominal voltage : for Zinc cathode = 1.5V to 1.65V ; for lithium cathode =
1.5 V to 3.0V
- End point Voltage : for Zinc cathode = 1.0V to 1.4V ; for lithium cathode =
1.2 V to 2.0V
Over excited synchronous motors having leading power factor are widely used for
improving power factor of those systems which employ a large number of induction
motors and other devices having lagging power factor such as welders and fluorescent
lights etc.
In constant speed applications, synchronous motors are well suited due to their high
efficiency and high speed, such as centrifugal pumps, belt driven reciprocating
compressors, blowers, line shafts, rubber and paper mill etc.
Voltage in long transmission lines varies greatly at the end when large inductive load
are present and when inductive load disconnected suddenly, the voltage tends to rise
considerably above its normal value because of line capacitance. Therefore, by
installing a synchronous motor with field regulator, the voltage rise can be controlled.
150
N 1/Φ ; By decreasing flux, speed can be increased and vice – versa. The
flux of DC motor can be changed by changing the field current with the help
of shunt field rheostat.
N (V-Ia Ra), The speed / armature current characteristics show that greater
the resistance in armature circuit , greater fall in speed.
Shunt field of motor is connected permanently to a fixed exciting voltage, but
armature is supplied with different voltages by connecting it across one of the
several different voltage by means of suitable switchgear.
Q.49 Why are starters used for starting 3-phase induction motors? (4)
151
- The phase displacement between primary & secondary voltages must be the same for all
transformers, which are to be connected for parallel operation.
- The Phase sequence must be the same.
- All 3-Φ transformers must have same construction either core or shell.
CLASS APPLICATIONS
(1) Variable torque, power output N2 . Fans, centrifugal pumps.
(2) Constant –torque power output N. Conveyors, stokers, reciprocating
Compressors, printing presses.
(3) Inverse torque power output rating constant. Machine Tools, lathes, boring mills,
drill, planers,.
The Multi-speed motors are of slip ring type used for hoist, conveyor and elevator.
Q.51 Highlight the role of following in the generation of power
(i) Cogeneration.
(ii) Diesel generator. (7 x 2 = 14)
152
Ans: Co-generation
The interconnection of hydro and thermal power generation plant or reuse of waste product of
first generator as a fuel of second generator or other machine :
The large power system comprising several power stations load centres interconnected to
form a single grid operation of such a grid is controlled from a load centre or load dispatch
centre. The national load control centre is linked with various regional despatching stations.
The regional load centres send commands from power stations to control room periodically
by telemetric data transmission system.
The automatic load frequency control in the control system of generator turbine governor
basically aim to maintain a constant frequency/speed as a primary control. But setting of the
governor for turbine is changed according to instruction of regional control centres.
Therefore, The input of turbine from governor gets automatically adjusted by primary load
frequency control and frequency is maintained. The governor setting is determined by
economy load dispatch instructions from regional load control centres. The total load
cogeneration control is achieved by:
(ii) Diesel Generator: These are used in medium power plants upto 25MW for industrial
and marine applications. The transportable diesel generator sets are for remote location, and
small power plants for small lawns /farms etc.
Generators are usually gear driven to speeds of 1000 to 1500 rpm for 50 Hz. Diesel engines
have a speed range of 1500 rpm to 2250 rpm (higher speed for gas turbine driven set)
The critical speed and torsional or natural frequencies are checked for diesel generator set to
avoid resonance. Brushless excitation is commonly used. The diesel generators are mounted
on same bed plate of prime mover.
Q.52(a) Explain the functions and basic requirements of a protective relay. (7)
(b) Compare the merits and demerits of overhead lines with an underground distribution
system. (7)
Functions:
The protective relaying senses the abnormal conditions as a part of the power system and
gives an alarm or isolate that part from the healthy system. The relays are compact, self-
153
Basic requirements
- Selectivity, discrimination: The protective relaying should select the faulty part of the
system and should isolate, as far as possible only faulty part from the remaining healthy
system.
- Speed, Time: It is the time between fault instant and closing of relay. A rapid contact
fault cleaning i.e. 0.07 second with 60 kA rms value of current, has no damage to the
system but if it is 7 sec, the bus bar will destroy complete. Therefore, relay time must be
minimum as much as possible (i.e. in millisecond.)
Stability : A quality of protective system by the virtue of which , the protective system
remains operating and stable under certain specified conditions such as system disturbances,
through faults, transient etc.
Reliability : The protective relaying should not fail to operate in the event of faults in
protected zone. The reliability of protective systems depends on diverse aspects such as
protective gear manufactures, Electricity Boards & Associates.
154
(b)
(A) Merit 1. Easy maintenance & 1. Less disturbances for other system
Repair 2. system looks neat and beautiful
2. Low cost of installation 3. Mostly used in distribution system .
3. Mostly used in (Medium and low voltages upto 11kW)
transmission system due to 4. Less lightening thunder effect
effective voltage upto
400 kV
4. Less skilled staff is
required.
(b) What is inductive interference? How is it caused and what steps are necessary to
reduce its effect? (7)
A welding process is required to join two metal parts by heating them to melting point.
Electrical welding is used for joining of fabricating structures, machinery parts, pipes, bus
bars, bridges, ships railway bogies etc.
- Resistance welding : In resistance welding the current is passed through the joint to be
welded and the heat is caused by I2 R losses in the joint, melting of metal and subsequent
welding under pressure between faces to be welded.
- Arc welding : The heat is produced by the arc struck between the welding electrode and the
metal to be welded. Pressure is not applied between faces to be welded.
155
Q.54 Identify the motor being used in the ceiling fan and explain the method of its control.
(14)
Single phase induction motor is used for ceiling fan; this motor is more or less similar
to polyphase induction motor, except that
(ii) Centrifugal switch is used in some types of motors in order to cut out a winding
used only for starting purposes; it has distributed stator winding and squirrel
cage rotor.
When single phase supply is fed, its stator winding produces a flux (or field) which is only
alternating i.e: one which alternates along one space axis only. It is not a synchronously
revolving (or rotating flux, as in the case of a two or three phase stator winding fed from 2 or
3 phase supply). Now, an alternating or pulsating flux acting on stationary squirrel cage rotor
can not produce rotation (only revolving flux can). Therefore, single phase is not self-starting.
To make single phase induction motor self starting the following method is adopted.
Split Phase method: In this method, phase is split by introducing a resistance and a switch
in starting winding and both windings are connected in parallel with supply.
156
157
158
(iii) The carrier current protection: The carrier current protection scheme is used for
protection of transmission lines. The information about short circuit/earth fault of the
power transmission line is conveyed through carrier communication link. The carrier
current of frequency range 30 to 200 KHZ or 80 to 500 KHZ and received through the
transmission lines for the purpose of protection. In this system, each end of the line is
provided with identical carrier current equipment consisting of a transmitter, receiver,
line tuning unit, master oscillator, power amplifier etc.
Q.56 How are the power angle characteristics of synchronous machine obtained. (8)
Ans: Power angle characteristics of a synchronous machine :
P = (VE/ X ) sinδ
Where;
P = Power
V= Terminal voltage
E= Induced e.m.f.
P = ± Vi Ef. sinδ / XS
δ is considered positive for generator mode and negative for motor mode.
159
The power output of a synchronous machine can be changed only by changing its
power output. Change in excitation does not change in power output. The change in
excitation will give only change in e.m.f and negative power supplied by the machine.
Ans:
Ia = (V-Eb)/Ra
160
When motor is at rest, no back emf will develop in armature. Now, if full supply
voltage is applied to stationary armature, it will draw very large current because
armature resistance is relatively very small.
∴ Ia = (V/Ra) ; Eb = 0
This excessive current is large and this will damage commutator, brushes etc. To
avoid this situation, a resistance is introduced in series with the armature for short
duration with the help of starter, which limit the starting current to safe value.
The starting resistance is gradually cut off and motor gets full speed with back e.m.f
under safe limit
Now Ia = (V-Eb )/Ra
However, very small motor can be started from rest by connecting directly to supply
These motors have relatively large armature resistance than larger motors. These
small motors also have low value of inertia.
161
Φm. The resultant flux Φr at any instant is given by the vector sum of the 3 individual
fluxes, Φ 1, Φ2, Φ3 due to 3 phases
Fig. (a)
φ1 = √3/2 φm
φ2 = - √3/2 φm
φ3 = 0
∴ φr = 3/2 φm
And θ = 120o
φ1 = √3/2 φm , φ2 = 0, φ3 = - √3/2 φm
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Mathematical proof :
Q.59 Derive the hystersis torque expression of the hystersis motor. (8)
Ans: Hystersis torque expression of the hystersis motor:
Usually, the shaded pole principle is employed for hystersis motor. The rotor of this
motor is smooth, having high retentively, so that, hystersis loss is high. It has no
winding, the rotor poles magnetically locked up with the revolving stator poles of
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opposite polarity. However, rotor poles always lag behind stator poles by a small
angle ‘ ’ and the rotor revolves synchronously.
The mechanical power developed by rotor is given by
Pm = Ph (1-S)/ S
Pm = Mech power developed
Ph = Hysteresis loss in rotor and hysteresis torque
Th = (9.55 Pm )/NS
Pm = (Eb Ia - losses)
= (60/2π ). (Pm/NS)
= (9.55 Pm / Ns)
This hysteresis torque is solely dependent on the area of rotor’s hysteresis loop.
• Generating stations
• Transmission systems
• Receiving station
• Distribution system
• Load points
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Generating stations:
In all these regions, there are switchgears. Bus bars are connecting bars to
which a number of local feeders are connected. Bus bars operate at constant
voltage. Besides the bus bars, there are other equipment in the electrical
schemes such as circuit breakers, CTs, PTs, etc. These equipments can be
installed according to various schemes depending upon requirements.
The power transformers are installed between two bus bars of different voltage
levels. A power transformer is the costliest, heaviest, and most important
equipment in substation.
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Receiving system & load points for a medium size industrial work:
The switch gear is installed in the substation of local points, such as industrial
works, railway substation, cinema house, large building, foundries etc. The
substation has following items:
- Transformer section
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The periodic rise and fall of water level of the sea can be used to convert wave energy
into electrical energy by moving turbine blades from fall and rise of water waves. In
about 24 hrs, there are two high & two low waves/ tides. The rise and fall of wave
follow sinusoidal curve.
The world first wave/dual power plant was commissioned at ‘Range’ in France. This
plant is of 240 mW capacity. There are three main components of a wave energy
power plant as follows:
- The dyke to form basic or basin
- Sluice ways from the basin to sea and vice-versa
- Power house
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The turbines, electric generators and other auxiliary equipments are the main
equipments of power house.
Advantage
- Tidal/wave power has a unique capacity to meet peak power demand effectively
when it works in combination with hydropower plant or thermal power
Disadvantage:
- Power transmission cost is high because the tidal/wave power plant are located away
from load centre.
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Closing time: The time interval between energizing of closing circuit and contact
touch with all poles.
Open close time: it is time between separation of arcing contacts all the poles and
contact touch in first pole deriving open close operation.
Reclosing time: Time between energizing of shunt trip release and contact toad in all
poles.
Close-open time: Time between contact touch in first pole and separation of arcing
contact in all poles during close open operation.
The rated characteristics of circuit breaker include rated normal current, rated voltage,
rated insulation level, rated transient recovery voltage, rated short circuit breaking
current, rated short circuit making current, rated operation sequence etc.
Q.62 Give a list of the factors involved for selecting a factory drive. (8)
Ans: Factors involved in selection of factory drives:
- Supply system : DC or AC
- Rating : kW, MW
- Drive : constant speed, Variable speed
- Special Condition : High starting torque, Hazardous location,
Traction duty etc.
- User’s requirement : Duty cycle, noise level, load characteristics etc
off load, load condition
- Enclosures & cooling requirement.
- Environment condition
- Applicable standards
- Cost considerations
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(ii) Copper losses : These losses are due to ohmic resistance of transformer windings.
Therefore, WC I12R1 + I22 R2 = I12 R01 or I22 R02
Q.64 Describe how the synchronous reactance of a synchronous machine is determined. (6)
Ans: Synchronous reactance determination of a synchronous machine
Eo = No load e.m.f.
E= On load induced e.m.f.
V= Terminal voltage
Z= (Ra2 + X L2) i.e: impedance
I= Armature current/phase
= Load p.f. angle
Xs = Synchronous reactance
Hence, Xs = XL + Xa
Q.65 Explain how speed control is achieved for DC shunt motors. (4)
Ans: Speed control for DC shunt motor
(i) Flux control method: N 1 /Φ, By decreasing flux, the speed can be increased and
vice-versa. The flux of d.c. motor can be changed by changing Ish with the help of
shunt field rheostat.
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N1 (V - Iao Ra)
Or N1 Eb1
and N2 ( V - I a2 Rt)
Rt = (R + Ra) Additional resistance added with armature in series
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N 2 E b2
∴ =
N 1 E b1
V − I a. R t
=
V − I ao R a
Ward Leonard System: This system is used for wide range of speed control (i.e:
10:1) and it will provide a very sensitive speed control as required in colliery winders,
electric excavators, elevators.
By applying variable voltage across its armature, any desired speed can be
obtained. This variable voltage is supplied by a motor-generator set which consists of
either d.c. or an a.c motor M2 directly coupled to the generator.
Q.66 Explain how the circuit model of an induction motor is obtained from no-load and
block-rotor tests. (4)
Ans: (a) Induction Motor model :
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The method is for finding Go and Bo in running motor synchronously so that slip S = 0. In
practice it is impossible for an induction motor to run at synchronous speed, due to friction
and windage losses. Therefore, induction motor runs at synchronous speed by another
machine, which supplies the friction and windage losses. In this case, the above circuit
behaves like an open circuit.
Therefore, s= 0
RL =
Hence current drawn by motor is I0 only.
Let V= applied voltage/phase
Io = Motor no load current /phase
W= Wattmeter reading i.e. input power in watt
Y0= exciting admittance of Motor
W= 3Go.V2 or Go = W/3V2 ; ( Go is no load conductance )
Wo = 3 . VL. Io .cos o
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Q.67 Draw the torque-speed characteristics of a single phase induction motor and explain
how it can be obtained. (6)
Ans: Torque-speed characteristics of a single phase induction motor
Torque developed by a single phase induction motor depends on its speed. But relationship
can not be represented by simple equation. It is easier to show the relationship between
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torque T and speed N in the form of curve. In this curve, it has been observed that at N=0, T1
= 1.5T and Tmax = 2.5T. This is called breakdown Torque.
At full load, the motor runs at a speed of N. when mechanical load increases, the motor speed
decreases till motor torque again becomes equal to the load torque. As long as the torques are
in balance, the motor will run at constant speed but lower speed. If the load torque exceeds to
2.5T, then motor will suddenly stop.
Q.69 (a) Describe the primary and back-up protection features that are provided for
transmission lines. (8)
(b) Describe a typical coreless type of induction furnace and its special features. (8)
Ans: a Primary and back-up protection:
The primary protection is essential protection provided for protecting a machine. As a
precautionary measure, an additional protection is generally provided and is called a
“Back up protection”. The primary protection is the first to act and back-up protection
is the next in the line of defence. Therefore, if primary protection fails, the back-up
protection comes into action and removes the faulty part from one healthy system.
When main protection is inoperative for purpose of maintenance, testing etc. The
back-up protection acts like main protection. The back up protection can be classified
as:
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- Relay back-up : Same circuit breaker is used by both main and back-up
protection, but protective systems are different. Separate trip coils may be
provided for the same-breaker.
- Breaker Back-up: Different breakers are provided for main and back up
protection, both the breakers are being in the same station.
- Remote back-up: Main & back up protection provided at different stations and
are completely independent.
- Centrally coordinated back-up: Here system is having central control and it can
be provided with centrally controlled back-up. The central coordinating station
receives information about the abnormal conditions through high frequency
carrier signals. The stored programme in digital computer determines the correct
switching operation as regard severity of faults, system stability etc. Main
protection at various stations and back-up protection for all stations is at central
control centre.
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- The relay responds to the vector difference between two i.e. I1~ I2 which includes
magnitude and/or phase angle difference.
Differential protection is generally a unit protection. The protected zone is exactly
determined by location of CTs. The vector difference is achieved by suitable
connection of CT or PT secondaries.
(ii) Inductive Interference in Transmission lines:
Inductive interference depends upon gradient of voltage at surface of conductor or
corona. The inductive reactance affected from conductor attenuation at 1000 kHz
varies from 0.3 to 0.9 dB/mtr.
(iii) Nickel cadmium cell
Nickel cadmium cells employ a solution of potassium hydroxide in distilled water as
electrolyte with a specific gravity of 1.200 at 16oC.
- Anode: NiOH and specially treated graphite
- Cathode: CdO, Fe2 O3
Cadmium oxide or iron oxide.
- AV. Voltage 1.2 V/Cell
1.3 V/Cell when fully charged.
The condition of battery can be determined with a voltmeter during charge or
discharge, while charging Ni-Cd battery, the temperature of electrolyte should not be
allowed to exceed 46oC. The battery does not bubble the gas until after the first
41/2 hrs when charging at the 7 Amp rate. No finishing rate is needed. The voltage of
charger should not be above 1.85 volt per cell.
Application : Type:
(1) Air craft, Emergency power application, Vent type
Industrial power supply communication
Equipment
Q.72 Give the lay-out of a typical thermal-power plant and briefly explain the working of
the super heater and the condenser. (8+4+4)
177
(i) Super heater: The steam produced inside boiler is nearly saturated. This steam as
such should not be used in the turbine because the dryness fraction of the steam-
leaving boiler will be low. This results in the presence of moisture, which causes
corrosion of turbine blades etc. Therefore, to rise the temperature of steam, super
heater is used. It consists of several tube circuits in parallel with one or more return
bends connected between headers. Super heater tube range is from 1 to 2 inch in
diameter.
(ii) Steam condenser: Steam condenser is required to receive the exhaust steam from the
turbine or engine to condense it and maintain a pressure of the exhaust lower than
atmospheric. Some extra work is obtained due to exhaust at pressure lower than the
atmospheric. This improves the efficiency of the plant. Air inside the condenser
should be pumped out continuously in order to maintain vacuum. The condensation
of steam occurs in the range of 250C to 380C.
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Advantages:
Q.73 With a neat diagram describe the construction of a lead acid cell and explain the
process of charging and discharging of the cell. (8+4+4)
Ans: Construction of lead acid cell and process of charging and discharging of cell:
A lead acid type batteries are known as secondary battery or cell, which can be
charged for reuse. This cell has following components:
- Separator : These are thin porous sheet suspended between positive and negative
electrode to prevent short circuit.
- Container : It contains all plates, electrolyte, separators, vent plug, cell connectors
and battery terminals. It is made of hard rubber.
- Cell connectors : These are used to connect a number of cells of same type (either
negative or positive) in series so as to provide required voltage.
Charging process:
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Discharging process:
When cell is fully charged anode is PbO2 and cathode is Pb. An external load is
connected to electrodes of cell, therefore, an electric current flows in load. The process is
known as discharging. A fully discharge cell has a voltage 1.8 volts and specific gravity
1.180.
The reaction-taking places at electrodes are
At anode : PbO2 + H2 + H2SO4 → PbSO4 + H2O
At Cathode: Pb + SO4→ PbSO4
Q.74 (a) What are the criteria for the classification of transmission lines as short, medium
and long lines? (8)
(b) Draw the schematic diagram of a directional overcurrent relay and explain its
working. (4)
(c) What are the advantages of high voltage transmission and its limitations. (4)
Ans: (a) Classification of transmission lines (Power)
The network of transmission lines is formed by three phase AC system. This is required
for Bulk power transfer from large group of generating stations to main transmission
network.
Short distance power transmission lines has voltage ≤ 11000 volts. These lines are
required to carry power from main sub-station to local distribution area. Further power
supply is distributed through pole mounted sub- station or plinth mounted sub-station.
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A long distance power transmission line has voltage range 220 kV to 750 kV. These lines
are used for transfer of power from sending end to receiving end or for system inter-
connection for exchange of power between independently controlled networks. For longer
transmission lines, higher power transmission voltages are necessary (p v2). Higher
voltage gives lesser current, lesser I2R losses, higher power transferability.
(b) : Schematic diagram of directional over current relay with its working:
Diagram:
The directional over current protection comprises over current relay and power
directional relay in a single relay casing. The power directional relay does not measure
the power but it is arranged to respond the direction of power flow. The directional
operation of relay is used where the selectivity can be achieved by directional relaying.
The directional relay recognizes the direction in which fault occur relative to the location
of relay. It is set such that it will actuate for fault occurring in one direction only. It does
not act for faults occurring in the other directions. In above diagram consider a feeder AB
phasing through subsection 1. The circuit breaker in feeder 1B is provided with a
directional relay R, which will trip the CB, if fault power flow in direction 1B alone.
Therefore, for faults in feeder 1A, the CB does not trip unnecessarily. However, for
faults in feeder 1B, CBB trips.
The choice of transmission systems and rated voltages for a transmission line is made
from HV AC (upto 220 kV) EHV AC (400 kV – 750 kV) UHVAC (above 760 kV AC)
depending upon technical and economic consideration.
Advantages :
- High power transferability of AC lines P V2.
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- Bulk power transfer from large group of generating stations upto main
transmission network.
Limitations:
In case of very long lines of above 500 km, intermediate switching sub-stations
are necessary to install the shunt reactors for compensation.
- Right of way:
- Line insulation:
The creep age distance (leakage distance) determined on the basis of required
impulse with stand level.
- Corona:
The critical value of voltage stress depends upon pressure, temperature, humidity,
pollution level in air.
182
183
full speed, a centrifugal switch S opens and cuts off both starting winding and capacitor from
the supply. Thus leaving only the running winding.
The torque developed by split phase motor is proportional to size of the angle between IS &
Im.
(ii) Switched Reluctance Motor :
This type of motor has either conventional split phase stator and a centrifugal switch for
cutting off auxiliary winding or a stator similar to that of permanent split capacitor run motor.
The stator produces resolving field. The working of this motor is that when a piece of
magnetic material is located in a magnetic field, a force acts on the material, tending to bring
it into the most dense portion of a field. The force tends to align the specimen of material in
such a way that the reluctance of the magnetic paths that lies through the material will be
minimum.
When stator winding is energized, the revolving magnetic field exerts reluctance
torque on the unsymmetrical rotor tending to align the salient pole axis of the rotor with the
axis of resolving magnetic field. If the reluctance torque is sufficient to start the motor and
its load, the rotor will pull into step with the revolving field and continue to run at speed of
revolving field. However, rotor poles lag behind stator poles by certain angle known as
torque angle.
The constant speed characteristic of a reluctance motor makes it very suitable for
signaling devices, recording instruments, timers, phonograph etc.
(iii) Stepper Motors
This motor rotate through a fixed angular step in response to each input current pulse
received by its controller. These motors can be controlled directly by computers,
microprocessors and programmable controllers. These motors used for precise positioning of
an object or precise control of speed without using closed loop feed back. The unique feature
of this motor is that its output shaft rotates in series of discrete angular interval or steps, one
step being taken each time a command pulse in received. When a definite number of pulses
are supplied, the shaft turns through a definite known angle. So that the motor is well suited
for open loop position control because no feed back is taken from output shaft.
184
185
Truth Table
A B C θ
+ 0 0 00
0 + 0 300
0 0 + 600
+ 0 0 900
These motors develop torques ranging from 1 MN upto 40 N-m in a motor of 15 cm diameter
suitable for machine tool applications. Their power output ranges from about
1 W to Max of 2500 W. The moving part in stepper motor is its rotor which has no winding,
commutator or brushes. Therefore, the motor is quite robust and reliable.
Q.76 What are the conditions for satisfying parallel operation of single phase transformer?
Deduce an expression for the load shared by the two transformers in parallel when
the transformers have equal voltage ratio. (8)
186
V = Secondary voltage
I1 = Input current
I2 = output current
v = voltage drop
v = I1 Z1 = I2Z2 = IZ12
Z12 = Z1 || Z2
I = equivalent current at output
∴ I2 = I Z1 / (Z1 + Z2)
∴ I1 = I Z2 / (Z1 + Z2)
187
Q.77 Explain two important functions served by the damper winding in a synchronous
motor. State the various applications of synchronous motor. (7)
- The oscillatory motion of the rotor about the operating point is considerably reduced
in amplitude and rotor quickly returns to the steady position.
- At starting, the rotor achieves a speed close to synchronous speed and the rotor and
stator fields lock into each other as soon as field excitation is switched on.
- It is capable of being operated under a wide range of power factors, both lagging and
leading. Hence, it can be used for power factor correction purpose.
- In textile industry high speed drives are necessary with wide range of control.
Variable frequency drives with synchronous motors are used for speed control.
- An artificial fibre plant needs variable frequency synchronous motor and are supplied
by static frequency converters.
Q.78 Explain the process of building up of voltage in d.c. shunt generator and give the
conditions to be satisfied for voltage built-up. (7)
Ans: The DC shunt generator runs at constant speed and some emf will be generated due to
residual magnetism in the main poles. This small e.m.f circulates a field current,
which produces an additional flux to reinforce the original residual flux. This process
will continue and generator builds up the normal generated voltage.
∴ E0 = iRf + L di/dt
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189
remains in circuit. The torque developed depends on angle between Im & Is, that is
approximate 800.
190
Q.80 Explain the principle of operation of two phase servo motor. Draw its torque-slip
characteristics. (8)
For low power (few hundred watts) control applications, the two phases
(balanced) servo motor is ideally switched. As it can be driven by means of a
relatively rugged (drift free) as an amplifier. The motor torque can be easily controlled
by varying the magnitude of a.c voltage applied to the control phase (Va) of the
motor, while second phase called reference phase (phase Vin) is excited at a fixed
voltage. Synchronous a.c voltage must be drawn from the same source. The control
phase is shifted in phase by 900 from the reference phase voltage by means of a phase
shifting networks included in voltage application stages of amplifier. The motor
torque gets reversed by phase reversal of control phase voltage.
For linear stable operation, the speed torque- characteristics of a servomotor must be
linear with negative slope (torque reducing with increasing speed). It is suitable for
the position control system.
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The steam after expansion through the turbine goes to condenser to improve efficiency of the
power plant by decreasing the exhaust pressure of the steam below atmosphere. Another
advantage of condenser is that condensed steam can be recovered and this provides a source
of good and pure feed water to the boiler. Thereby, reducing considerably the water softening
plant capacity. The air and non condensable gases are removed from the steam, when it
passes through the condenser.
Once fission process is started in a few nuclei of fission materials, it should be able to
continue throughout the remaining material without external interference. For each act of
fission, nearly 2.5 neutrons capable of causing further fission are proceeded. A self-sustaining
chain reaction is not possible. The leakage of neutron through it is inevitable. Both fission
and non fission reduces the number of neutrons actually available to maintain the chain
reaction.
The reproduction factor (k) is defined as:
k=
The desirable requirement of power reactor is that the system should be critical i.e
k =1. When we want to control reactor, we have to change the value of k. At the
time of starting the reactor value of k has to be raised above 1. This will increase the power
level.
Once the required power level has been reached, k is to be reduced to 1 and should
be kept at this value as long as the out-put rate is to be kept constant. When the load on the
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reactor is to be lowered, the k has to be reduced to slightly less than 1 till the required
power level has been reached. At this point k has to be brought back to 1. Similarly, while
shifting down the reactor, the k has to be reduced to less than 1, the chain reaction will die
down.
Q.82 Describe the construction, principle of operation and advantages of SF6 circuit
breaker. (8)
In general, the circuit breakers consist of two main parts, the poles and the mechanism. The
poles consist of contact and arc-extinguishing devices. The mechanism is the part to open or
close the contacts in the poles at the same time instantaneously (with max. 5 millisec.
Tolerance). The closing and opening procedures are performed through springs which are
charged by a servomotor and a driving lever. In the system, the closing springs are first
charged. If "close" button is pressed, the opening springs get charged while the contacts get
closed. Thus, circuit breaker will be ready for opening. The mechanical operating cycle of the
circuit breaker is (OPEN-3 Min CLOSE/OPEN-3 Min- CLOSE/OPEN) or (OPEN-0.3 sec-
CLOSE/OPEN-3 Min CLOSE/OPEN). The second cycle is valid when the circuit breaker is
used with re-closing relay. In that case, after the closing operation, the closing springs are
charged by the driving lever or by driving motor (if equipped). Thus, the circuit breaker will
be ready for opening and re-closing.
When manual or motor-drive is used, the circuit breaker will be ready to close. The closure
can be actuated by pressing the closing button located on the circuit breaker. It is
recommended to close it using remote control system for secure operations. The opening can
be performed either by opening button or remote controlled opening coil. In case of a fault,
the relay signal actuates the opening coil and circuit breaker opens. (This is mechanically a
primary protection system). In addition, there is an anti-pumping relay for preventing the re-
closing and opening of the circuit breaker more than one cycle (O - C - O) and for preventing
possible troubles created by remote closing button.
193
Applications:
194
- Since cost of dielectric heating is very high, so it is employed where other methods
are not applicable.
- For baking of sand cores, which are used in the moulding process.
- For drying of tobacco after glycerine has been mixed with in making cigarettes.
- The baking of biscuits and cakes etc. in the bakeries with the help of automatic
machine.
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The average output voltage of a thyristor controlled rectification can be changed with
the help of a thyristor chopper circuit, which can be made to interrupt dc supply at
different rate to give different average value of dc supply. The dc voltage so obtained
can be thus chopped with the help of thyristor chopper circuit.
Since the thyristor can be switched ON- OFF very rapidly, they are used to interrupt a
dc supply at regular frequency in order to produce an average (mean) dc supply
voltage. This is achieved from low level dc voltage to high level dc voltage. The
thyristor T1 is used for dc chopping and resistance ‘R’. The capacitance ‘C’ &
Thyristor ‘T2’ are used for commutation purposes. When T1 is fixed into conduction
by its control circuit, the current is set up through the load and common capacitance
‘C’ gets charged Via ‘R’ with polarity during ‘ON’ period. When T1 is ‘OFF’, second
thyristor ‘T2’ is triggered into conduction allowing ‘C’ to discharge through it (since
it acts as short circuit while conducting) which is reversed biased, T1 thus turns it
OFF. The discharge from ‘C’ leaves T2 with reverse polarity, so it is turned OFF,
where as T1 is triggered into conduction again.
It is seen that output voltage is present only when T1 is ON and OFF. The mean value
of output dc voltage depends on the chopped dc output of TON & TOFF..
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A differential relay is defined as the relay that operates when the phaser
difference of two or more similar electrical quantities exceed by a predetermined
amount. Suppose that current ‘I’ flows through the primary circuit to external fault. If
the two CTs have the same ratio and are connected as shown, no current will flow
through the relay and it will remain un-operational.
If now an internal fault occurs and if current flows to the fault from both sides, the
current flow through the relay will be I1 & I2. It may be noted that the fault current
need not necessary flow to the fault from both sides to cause current flow in the relay.
There may be flow of one side only or even same current flowing out of one side,
while a large current entering the other side will cause a different current.
This type of differential relay is likely to operate inaccurately with heavy load
through faults (external load). suppose identical CTs may not have identical
secondary current due to constructional feature and errors or under severe fault
condition CTs may saturate and cause unequal secondary currents and its difference
of secondary currents may approach the pick-up value of relay. This disadvantage is
overcome in percentage differential relay.
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The simple element of a transformer consists of two coils having mutual inductance and a
laminated steel core. The two coils are insulated from each other with steel core. Other
necessary parts are : some suitable container for assembled core windings from its container:
Suitable bushings (either of porcelain, oil- filled or capacitor type) for insulating and bringing
out the terminals from the tank. Core is constructed of transformer sheet steel laminations,
assembled to provide a continuous magnetic path with a minimum of air gap included. Steel
used is of high silicon content, sometimes heat treated to produce a high permeability and low
hysterisis loss at the usual operating flux-densities. Lamination is of varnish or an oxide
layer, which reduces eddy current loss. Thickness of lamination varies with frequency
inversely.
Transformers are of 2 general types distinguished by the manner in which primary and
secondary coils are placed around the laminated core :
Working principle :
A transformer is a static piece of apparatus by means of which electric power in one circuit is
transformed into electric power of the same frequency in other circuit. It can raise or lower
the voltage in a circuit but with a corresponding decrease or increase in current. The physical
basis of transformer is mutual induction between two circuits linked by a common magnetic
flux. It consists of two inductive coils which are electrically and magnetically linked through
a path of low reluctance. The 2 coils possess high mutual inductance. If one coil is connected
to a source of alternating voltage , an alternating flux is set-up in laminated core, most of
which is linked with other coil in which it produces mutually induced emf (according to
Faraday’s law of EMI e = MdI/dt). If the second coil circuit is closed, a current flows in it
and so electric energy is transferred (entirely magnetically) from the first coil to the second
coil. The first coil in which electric energy is fed from A.C. supply mains is called primary
winding and other from which energy is drawn out, is called secondary winding.
198
Q.85 What are the two types of constructions that are employed in synchronous machines?
Explain the two machines and give with reasons which of them are simple to model
and analyze. (2+4+2)
Ans: Two types of construction employed in synchronous machine are salient pole and
smooth cylindrical rotor type.
(i) Salient pole type or Projecting Pole Type:
It is used in low and medium speed (engine driven) alternators. It has a large
number of projecting (Salient) poles, having their cores bolted or dove tailed
onto a heavy magnetic steel of cast iron or steel of good magnetic quality.
Such generators are characterized by their large diameters and short axial
lengths. The poles and pole shoes (which cover 2/3 of pole–pitch) are
laminated to minimize heating due to eddy currents.
(ii) Smooth Cylinder Type:
Used for steam turbine driven alternators i.e. turbo alternators which run at
very high speeds. Rotor here is deigned mostly for 2-pole (or 4 pole) turbo
generators running at 3600rpm (or1800 rpm).
The central polar area is surrounded by the field windings placed in slots and
the field coils are so arranged around these polar areas that flux density is
maximum on the central polar line and gradually falls away on either side. It
should be noted that poles are non-salient i.e., they do not project out from the
surface of rotor. To avoid excessive peripheral velocity, such rotors have very
small diameters (about 1m or so). Hence, turbo generators are characterized by
small diameters and very long axial (or rotor) length. The cylindrical
construction of rotor gives letter balance and quieter-operation and less
windage losses.
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Cylindrical types are much easier to analyze than salient pole types because
they have uniform air gap, whereas in salient pole type air gap is much greater
between the pole than along the poles. Fortunately cylindrical rotor theory is
reasonably accurate in predicting the steady state performance of salient pole
type. Hence salient pole theory is required only when very high degree of
accuracy is needed or when problems concerning transients or power system
stability are not handled.
Q.86 Explain the characteristics of DC motors. Also give their applications. (6+3)
Ans: The characteristic curves of a motor are those curves which show relationships
between the following quantities:
1. Torque and armature current i.e Ta / Ia characteristic. It is known as
electrical characteristic.
2. Speed and armature current i.e N / Ia characteristic.
3. Speed and Torque i.e N / Ta characteristic. It is known as mechanical
characteristic.
Characteristics of series motor :
i. Ta / Ia characteristic:
Ta Φ Ia
Φ Ia (up to the point of magnetic saturation.)
Hence, before saturation
Ta Φ Ia and therefore Ta Ia2
As Ia increases, Ta increases as the square of the current. Hence , Ta / Ia
curve is a parabola before saturation. After saturation characteristic,
becomes straight line because Ta Ia (for Φ is independent of Ia)
Ta
ii. N / Ia characteristic:
N Eb / Φ
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iii. N / Ta characteristic :
i. Ta / Ia characteristic:
Ta Ia
ii. N / Ia characteristic:
iii. N / Ta characteristic can be deduced from (i) and (ii) above and is
shown in Fig :
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Current Ia
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Applications of DC motors:
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autotransformers are cut out and full supply voltage applied across motor. The
switch taking these adjustments from start to run may be air break (for small
motors) or oil immersed (for large motors) to reduce sparking.
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current. Hence such motors can be started with load. This additional resistance
is for starting purpose only. It is generally cut out as motor comes up to speed.
Q.88 What are the two advantages of stepper motors? Give a few applications of them &
explain permanent magnet stepper motors. (2+2+4)
Applications:
i) Operation control:
Others:
It has wound stator poles and permanently magnetized rotor poles. It has a
cylindrical rotor. Its direction of rotation depends on the polarity of stator current.
The stator has projecting poles. Rotor has 2 poles whereas stator has 4 poles and
the two stator poles are energized by one winding, the motor has two windings or
phases.
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Working :
When a particular stator phase is energized, the rotor magnetic poles move into
alignment with excited stator poles. The stator wdgs. A and B can be excited with
either polarity current (A+ refers to the +ve current i+A in phase A and A- to
negative current (i-A). Fig, shows condition when phase A is excited with the
current i+A. Here = 0o. If excitation is now switched to phase B, the rotor rotates
by a full step of 90o clockwise direction. When phase A is excited with –ve
current i-A, the rotor turns through another 90o in clockwise direction. Similarly
excitation of phase B with i+B further turns rotor through another 90o in same
direction. After this excitation of phase A with i+A makes rotor run through one
complete revolution of 360o.
It will be noted that in a permanent magnet stepper motor, the direction of rotation
depends on the polarity of phase current as below:-
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Q.89 (a) Explain the three main blocks of a solid state relay. (8)
(b) What is meant by grading of cables? Explain the two methods of grading. (8)
Ans: Solid-state relays
A solid-state relay conducts load current through one or more power transistors or
thyristors.
There are three main types of SSRs, classified by the type of input.
Reed-relay coupled SSRs send the input signal through the coil of a reed relay.
Closure of the reed switch triggers the thyristor into conduction.
Transformer-coupled SSRs send an ac control signal through the primary of a small
transformer. (In the case of dc control signals, the input first goes through a DC/AC
converter before hitting the transformer primary). Voltage from the transformer
secondary triggers the thyristor.
Finally, photo coupled SSRs isolate the input from the output through a
photosensitive diode or transistor. A control current actuates a light source which in
turn, lets a photosensitive semiconductor conduct trigger-current to actuate the power
device.
(b) The method of equalizing the stress in the dielectric of the cable is called the
grading of cables.
Methods of grading.
(i) Capacitance grading
(ii) Intersheath grading
Capacitance grading is provided within gas insulated lightning arresters containing
stacked zinc oxide varistors by means of a grading ring electrically connected to the
line terminal or, for arresters of the higher voltage ratings, by means of a plurality of
telescoping external electrostatic shields. The shields are arranged so that the degree
of overlap between sequential shields decreases from the line end to the ground end of
the varistor stacks. The capacitance grading is provided by the degree of overlap
between the sequential shields and the ratio of the radii of the overlapping shields.
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Intersheath Grading
In this method of grading, the same insulating material is used throughout the cable,
but is divided into two or more layers by means of cylindrical screens or intersheaths .
These intersheaths are connected to tappings from the supply transformer, and the
potentials are maintained at such values that each layer of insulation takes its proper
share of the total voltage. The intersheaths are relatively flimsy, and are meant to
carry only the charging current. Since there is a definite potential difference between
the inner and outer radii of each sheath, we can treat each section separately as a
single core cable.
Q.90 (a) Explain the features of a nuclear power plant with a suitable diagram. (6)
(b) Explain the term “cogeneration” and give two possible ways of cogeneration. (2+2)
(c) Write a short note on solar energy. (6)
Ans: Nuclear power station :
In the nuclear power station, energy is produced by nuclear fission of Uranium (U235),
Thorium (Th232), Plutonium (Pu239). The fission process takes place in a nuclear
reactor and material in certain circumstances can be made to become unstable and
transform themselves into ordinary chemical elements, iron nickel, silicon and
calcium and in this process set free a considerable amount of heat energy. Heat thus
generated will be taken away by fluid such as molten bismuth or liquid sodium. This
hot molten metal will convert feed water into high-pressure steam in a heat exchanger.
The steam thus generated is utilized to drive steam turbine coupled to an alternator
and exciter, thereby generating electrical energy.
Low temperature and low pressure steam from turbine is condensed in a condenser
with the help of circulated cooling water. From the condenser, water is again fed to
heat exchanger with the help of a feed water pump for converting it into steam.
In this power station transportation of fuel is not a problem, because one kg of atomic
fuel can generate as much electrical energy as can be generated from about
2500 tonnes of high grade coal.
The most suitable areas in our country for atomic power stations are western UP,
Rajasthan, Punjab and Haryana
Few examples of atomic or nuclear power stations are as follows:
1. Tarapur Atomic Power Stations, Tarapur near Mumbai.
2. Ranapratap Atomic Power Stations, Kota (Rajasthan).
3. Madras Atomic Power Stations, Kalpakkam (Tamil Nadu).
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(b) Co-generation
Cogeneration is a highly efficient means of generating heat and electric power at the
same time from the same energy source. Displacing fossil fuel combustion with heat
that would normally be wasted in the process of power generation, it reaches
efficiencies that can triple, or even quadruple conventional power generation.
Cogeneration equipment can be fired by fuels other than natural gas. There are
installations in operation that use wood, agricultural waste, peat moss, and a wide
variety of other fuels, depending on local availability.
The environmental implications of cogeneration stem not just from its inherent
efficiency, but also from its decentralized character. Because it is impractical to
transport heat over any distance, cogeneration equipment must be located physically
close to its heat user. Cogeneration plants tend to be built smaller, and owned and
operated by smaller and more localized companies than simple cycle power plants. As
a general rule, they are also built closer to populated areas, which causes them to be
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(c) Solar energy is the utilization of the radiant energy from the Sun. Solar power is often
used interchangeably with solar energy but refers more specifically to the conversion
of sunlight into electricity, either by photovoltaic and concentrating solar thermal
devices, or by one of several experimental technologies such as thermoelectric
converters, solar chimneys and solar ponds.
Solar energy and shading are important considerations in building design. Thermal
mass is used to conserve the heat that sunshine delivers to all buildings. Daylighting
techniques optimize the use of light in buildings. Solar water heaters heat swimming
pools and provide domestic hot water. In agriculture, greenhouses expand growing
seasons and pumps powered by solar cells (also known as photovoltaics) provide
water for grazing animals. Evaporation ponds are used to harvest salt and clean waste
streams of contaminants.
Solar distillation and disinfection techniques produce potable water for millions of
people worldwide. Simple applications include clotheslines and solar cookers which
concentrate sunlight for cooking, drying and pasteurization. More sophisticated
concentrating technologies magnify the rays of the Sun for high-temperature material
testing, metal smelting and industrial chemical production. A range of prototype solar
vehicles provide ground, air and sea transportation.
Q.91(a) What are the advantages of electrically produced heat? Explain the various types of
electric heating with their applications. (2+6)
(b) Explain the salient features of electrical propulsion. (8)
Ans: (a) Electrically produced heat possesses the following advantages over other forms of
Heat:
i. Cleanliness: The complete absence of dust and ash keeps cleaning cost to a
minimum.
ii. High efficiency of utilization : Practically 75% to 100% 0f heat produced can be
utilized.
iv. Ease of control : Simple and accurate control of temperature can be provided.
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Applications:
• Seaming and welding in manufacture of synthetics.
• In wood processing industry.
• For baking foundry cores.
• For food processing.
iv. Arc Heating :
If an air gap is subjected to a very high voltage , the air in between gets ionized due to
electrostatic forces. The ionized air is the conducting material, therefore, the current
starts flowing through the air gap in the form of continuous spark or arc. This arc is
produced in the arc furnaces by having air gap and electrode. With graphite or carbon
electrode the temperature obtained from the arc is between 3000oC and 3500oC.
There are two types of arc furnaces.
• Direct arc furnace.
• Indirect arc furnace.
Applications:
• In production of steel direct arc furnace is used.
• Submerged arc furnace is used for ferrous alloy manufacturer.
• Indirect arc furnace are used for melting of non ferrous metals, it is also used in
foundries where small quantities of metal is required intermittently.
(b) Salient features of electrical propulsion.
Electric propulsion is a form of space craft propulsion used in outer space. This type
of reaction utilizes electric energy to obtain thrust from propellant carried with the
vehicle.
Types of electric propulsion :
The various technologies of electric propulsion are usually grouped in three types
based on the type of forces used to accelerate the ions of the plasma.
Electrostatic: If the acceleration is caused mainly by the coulomb forces the device is
considered electrostatic.
i. Electrothermal: In this type, electromagnetic fields are used to generate a plasma
to increase the heat of the bulk propellant. The thermal energy imparted to the
propellant gas is then converted into kinetic energy by a nozzle of either physical
material construction or by magnetic means.
ii. Electromagnetic: In this ions are accelerated either by the effect of an
electromagnetic fields where the electric field is not in the direction of the
acceleration.
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Advantages of Electric Propulsion used in the battery electric vehicle that uses
chemical energy stored in rechargeable battery packs.
As Electric vehicle, it employs electric motors and motor controllers instead of
internal combustion engine
The concept of baterry electric vehicle is to charge batteries on board of vehicles
for propulsion using the electric grid.
1. No pollutants are emitted directly by the vehicle potentially reducing urban pollution.
2. Gasoline is indirectly replaced by whatever is being used to generate domestic
electricity, reducing dependence on foreign commodities. The electrical energy
stored within the battery can be generated by any source, including renewable,
nuclear, natural gas, coal and petroleum.
Y- Connection:
Used at substation end of transmission line where voltage is stepped down. There
is 30o shift between primary and secondary line voltages which means Y- can be
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Q.93 Explain the constructional features of synchronous generator. What are the two types
of generators? Derive emf equation of a synchronous machine. (10)
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having alternate N and S poles fixed to its outer rim. The magnetic poles are excited
from direct current supplied by a d.c source which is belted or mounted on the shaft of
the alternator itself. Because the field magnets are rotating, this current is supplied
through two slip rings. As the exciting voltage is relatively small, the slip ring and
brushes gear are of light construction.
Types of synchronous generator :
There are two types of rotor used in synchronous generator. Accordingly, A
synchronous generator may be classified as:
(i) Salient pole type
(ii) smooth cylindrical type.
Salient pole type has a large number of projecting poles having their cores bolted or
dovetailed on to a heavy magnetic wheel of cast iron or steel of good magnetic
quality. Such generators are characterized by their large diameter and short axial
length.
Smooth cylindrical type runs at very high speed. The rotor consists of a smooth solid
forged steel cylinder having a number of slots milled out at intervals along the outer
periphery for accommodating field coils. Such rotors are designed mostly for 2 pole
turbo generator running at 3600 rpm. These are characterized by small diameter and
long axial rotor length. The cylindrical construction of the rotor gives better balance
and quieter operation and less windage losses.
E.M.F equation of synchronous machine :
Let Z = no. of conductor or coil sides in series / phase.
= 2T – where T is the no of coils or turns/phase.
P = No. of poles.
F = frequency of induced e.m.f in Hz.
Φ = flux /pole in webers
Kd = distribution factor = sin m /2 / msin /2
KC & KP = Pitch or coil span factor = cos /2
Kf = form factor = 1.11 – if e.m.f is assumed sinusoidal
N = rotative speed of the rotor in r.p.m
In one revolution of the rotor (i.e in 60/N seconds) each stator conductor is cut by a flux of
Φ p webers
dΦ = Φ p and dt = 60/N seconds
∴ average e.m.f induced per conductor
= dΦ/ dt = Φ p / (60/N) = Φ Np / (60) volts
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Q.94 (a) Give comparison between squirrel cage and slip ring induction machine? Discuss
the working principle of three phase induction motor. (8)
(b) State different methods of speed control of three phase induction motor. Explain
any one of the method in detail. Also draw torque-speed characteristics. (8)
Almost 90% of induction machines are Squirrel-cage type, because this type of rotor
has the simplest and most rugged construction imaginable and is almost
indestructible. Rotor is of cylindrical type laminated core.
Rotor slots are skewed for (a) quiet running by reducing magnetic hum. (b) helps in
reducing locking tendency of the rotor i.e. tendency of rotor teeth to remain under
stator teeth due to direct magnetic attraction.
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But under normal running conditions wound rotor is short circuited on itself just like
squirrel case rotor.
In A.C. motors, rotor does not receive electric power by conduction but by induction
in exactly the same way as the secondary of 2-wdg transformer receives its power
from the primary. So such motors are known as induction motors.
The principle of a 3Ø 2 pole stator having 3 identical wdg placed 120o apart. The flux
(assumed sinusoidal) due to 3 Ø wdg is shown. Let Ø m be maximum value of flux
due to any one of the 3Ø . The resultant flux Ør at any instant, is given by vector sum
of individual fluxes Ø 1 Ø 2 Ø 3 due to 3 Ø. Let us consider values of Ør at four
instants 1/6th time period apart corresponding to points marked 0,1, 2 and 3.
Here Ø1 = - 3 Øm
2
Ø2 = 3 Øm,
2
Ø3 = 0
Here Ø1 = - 3 Øm; Ø2 = 0, Ø3 =- 3 Øm
2 2
Ør = 3
2
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Ø1 = 0, Ø2 = 3 Øm; Ø3 = - 3 Øm
2 2
Hence
1. Resultant flux is 3 Øm i.e. 1.5 times the maximum value of flux
2
due to any phase.
2. Resultant flux rotates around stator at synchronous speed given by
Ns = 120 f/p
(b) Different methods of speed control of three phase induction motor
The speed of an induction motor is given by
N = (f/p) x (1-S)
Where, N is speed
F is frequency
P is the no of poles.
S is the slip.
Thus speed of three phase induction motor is controlled by the following methods:
a. Rheostatic control
b. pole control
c. cascade control
e. Frequency control
Pole changing method: Winding on the stator may be so arranged that with different
connection, different numbers of poles are available which give different speeds. The
choice of the number of poles on the pole changing winding is in the ratio 2:1,
alternatively independent winding may be provided on the rotor.
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Q.95 Describe the construction of hystersis motor and show that it builds a running torque
both at synchronous and asynchronous speed of the rotor. (8)
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This motor builds a running torque both at synchronous and asynchronous speed of
the rotor It accelerates rapidly, changing from rest to full speed almost
instantaneously.
Q.96 (a) What do you understand by the term cogeneration? Give its significance. (8)
Ans: (a) Cogeneration is a high efficiency energy system that produces both electricity
(or mechanical energy) and valuable heat from single fuel source. Cogeneration is
some times known as combined heat and not only generates power but also provides
heat for industrial purposes. For example a hospital cogeneration plant would
produce some of power and all hot water needed for its laundry and hot water system
from the waste heat it generates.
Significance :
It reduces energy costs and green house gas emissions typically by upto two-third. In
addition to reduction in cost, cogeneration also increases resource utilization. It offers
major economic and environmental facilities because it turns otherwise wasted heat
into a useful energy source.
Cogeneration is a proven and reliable technology currently operating at over 100 sites
across Australia.
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There is significant potential for cogeneration plants fired by other fuels, including
biomass (e.g., plant waste from sugar or cotton harvesting), or biogas (e.g., methane
produced by sewage works or piggeries).
(b)
(i) Wind
Wind energy used in wind mills to pump water from wells and low level to high level.
Windmills do not generate electricity but are used to grind grains with large grinding
stones. Today wind is used to make electricity, blowing wind spins the blades on a
wind turbine, which is called wind turbine. The blade of turbine is attached to a hub
i.e. mounted on a turning shaft. The shaft goes through a gear transmission box where
the turning speed is increased. The transmission is attached to a high-speed shaft,
which turn generator that makes electricity. If wind gets too high, turbine has a brake
that will keep blades from turning to fast and being damaged. For efficient working,
wind speeds must be above 12 to 14 miles/hr., to turn the turbines fast enough to
generate electricity. Turbines usually produce 50 to 500 kW of electricity each.
(ii) Wave
Waves are caused by action of winds on sea. Waves can be many meters high and
contain a great deal of energy, which can be harnessed to drive turbines that generate
electricity.
Wave energy can be harnessed in coastal areas, close to shore. There has been one
such device working on the island of Islay in Scotland since early 1990s producing
75 kW of electricity. Wave energy collectors are of 2 main types :
(a) First type directs waves into man-made channels, where the water passes
through a turbine that generates electricity.
(b) Second type uses up and down movement of wave to push air.
Advantages : Non-polluting and don’t affect wild life.
Disadvantages : Turbines can be unsightly and no constant supply of energy. Wave
heights vary.
(iii) Bio fuel:
Bio fuel are transportation fuel like ethanol and bio-diesel that are made from biomass
materials. These are usually blended with petroleum fuels-gasoline and diesel fuel,
but they can also be used on their own. Ethanol and bio-diesel are expensive than
fossil fuels that they replace but are cleaner fuels, producing fewer air pollutants.
Ethanol is an alcohol fuel made from sugars found in grains, such as corn, sorghum
and wheat as well as potato skins, rice etc. Most of ethanol in US is distilled from
corn.
Bio-diesel is a fuel made with vegetable oils, fats or greases such as recycled
restaurant grease. Bio-diesel fuels can be used in diesel engine without changing
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