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AC MachinesUNIT 5 AC MACHINES
Structure 5.1 Introduction
Objectives
5.2 Transformer 5.2.1 Working Principle of a Transformer
5.2.2 Equivalent Circuit
5.2.3 Open Circuit Test and Short Circuit Test
5.3 Induction Motor 5.3.1 Construction Details
5.3.2 Rotating Magnetic Field
5.3.3 Principle of Operation of a Three Phase Induction Motor
5.3.4 Phasor Diagram of an Induction Motor
5.3.5 Equivalent Circuit
5.3.6 Air Gap Power, Torque and Output Power
5.3.7 Torque-slip (T-S) Characteristics
5.3.8 Starting of Induction Motor
5.3.9 Speed Control of Three Phase Induction Motors
5.3.10 Applications of Induction Motors
5.4 Single Phase and Special Purpose Motors 5.4.1 Single Phase Induction Motors
5.4.2 The Universal Motor
5.4.3 Permanent Magnet DC Motor (PMDC)
5.4.4 AC Servo Motor
5.4.5 DC Servo Motor
5.5 Summary 5.6 Answers to SAQs
5.1 INTRODUCTION
In the previous units we have studied the basic fundamentals of the AC and DC circuits. We have got familiar with electro-magnetism also. In this unit we shall study the transformer and AC machines. In the AC machines we shall focus on the basic principles and construction of the transformer, induction motor and single-phase motors. The transformer is one of the most important component of a variety of electrical circuits ranging from low power, low current electronic and control circuits to ultra high-voltage power systems. The induction machine finds wide applicability as a motor in industry and in its single-phase form in several domestic applications. More than 85% of industrial motors in use today are infact induction motors.
Objectives After studying this unit, you should be able to
• describe the transformer and its basic working,
• explain the testing of the transformer,
• describe the basic construction of three phase induction machine,
• explain the working principle and speed-torque characteristics of a three phase induction motor,
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Electrical
• understand the variety of single phase induction motors, and
• explain the servomotors and permanent magnet DC motor.
5.2 TRANSFORMER
Transformers are used widely in AC power systems because they make possible power generation at the most economical level (10-20 kV), power transmission at an economical transmission voltage (400-1000 kV) and power utilisation at most convenient distribution voltages (230/400 V).
Transformers are used for :
(a) Changing voltage and current levels in electric power systems.
(b) Matching source and load impedances for maximum power transfer in electronic and control circuits.
(c) Electrical isolation (isolating one circuit from another).
As there is no rotating parts, the construction of transformer is comparatively simple. Transformer is a machine with highest efficiency amongst the electrical machines.
The following are the main parts of a transformer :
(a) Magnetic circuit mainly comprising of limbs and yoke.
(b) Electric circuit mainly comprising of windings and insulations.
(c) Tank to accommodate the whole assembly, i.e. oil, cooling devices, conservators, etc.
We shall define the transformer as :
“A transformer is a static device that consists of two or more than two coils, wound on an air core or a magnetic material core, that are electrically isolated but magnetically coupled. It transforms one or more electrical parameters such as voltage, current, impedance and so on from one alternating circuit to another.”
Figure 5.1 : A Transformer with N1 and N2 Number of Turns in Primary and Secondary Respectively N1 ≠ N2 (Normally)
A transformer, in its simplest form consists essentially of two insulated windings interlinked by a common or mutual magnetic field established in a core of magnetic material. A winding connected to the AC supply is termed as primary winding and other winding supplies the load is termed as secondary winding.
5.2.1 Working Principle of a Transformer When the primary is connected to the AC voltage source, an alternating flux is produced in the core with an amplitude depending on the primary voltage, frequency and number of turns. This mutual flux links the other winding, called the secondary. A voltage is induced in this secondary of the same frequency as the primary voltage but its magnitude depends on the number of secondary turns.
The voltage induced in the primary is
E1 = 4.44 φ f N1 volts . . . (5.1)
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AC Machineswhere E1 = Voltage induced in primary, V
φ = Flux in the core, Wb, and
N1 = Number of turns in primary winding.
Similarly, the voltage induced in the secondary is
E2 = 4.44 φ f N2 volts . . . (5.2)
where E2 = Voltage induced in secondary, V, and
N2 = Number of turns in secondary winding.
Thus from the Eqs. (5.1) and (5.2), we have
=1 1
2 2
E NE N
. . . (5.3)
If the secondary voltage is greater than the primary voltage, the transformer is called a step-up transformer (N2 > N1). If it is less, it is known as a step-down transformer (N1 > N2). If N1 = N2, the transformer is said to have a one to one ratio.
The ratio of transformation may be given by
1 1
2 2
E NaE N
= = . . . (5.4)
As in the case of transformers,
Input volt amperes (VA) = Output VA
1 1 2 2V I V I= . . . (5.5)
where V1 and V2 are the terminal voltages of the primary and secondary windings respectively.
Also and as winding resistances and reactances are very small. From the Eq. (5.4), we can get
1V E≈ 1 22V E≈
1 1 1
2 2 2= = = = 2
1
E V N IaE V N I
. . . (5.6)
SAQ 1 (a) What is the need of a transformer?
(b) Explain the working principle of a transformer.
(c) What do you mean by step up and step down transformer?
(d) List the main components of a transformer.
5.2.2 Equivalent Circuit We can now develop the equivalent circuit of the transformer. The transformer windings have some resistances and reactances. We can visualise the primary winding current to comprise two components as below.
(a) Exciting current 0I whose magnetising component mI creates mutual flux φ and whose core loss component iI provides the loss associated with alternation of flux.
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Electrical
(b) A load component 2I ′ which counter balances the secondary magnetomotive force (mmf) 2 2I N so that the mutual flux remains constant independent of load.
Thus, primary current 1I may be represented by
1 0I I I2′= + . . . (5.7)
where 2 2
2 1
I NI N′= . . . (5.8)
We shall represent the primary winding/side quantities by subscript ‘1’ and secondary by subscript ‘2’. Thus, R1 and X1 are resistance and reactance of primary winding respectively and R2 and X2 are resistance and reactance of secondary winding respectively. We can represent the transformers equivalent circuit as below shown in Figure 5.2.
Figure 5.2 : Equivalent Circuit of a Transformer
We can also develop the equivalent circuit referred to either primary side or secondary.
Figure 5.3 : Equivalent Circuit of a Transformer Referred to Primary Side
where 2
12 2
2
NX XN
⎛ ⎞′ = ⎜ ⎟
⎝ ⎠ . . . (5.9)
2
12 2
2
NR RN
⎛ ⎞′ = ⎜ ⎟
⎝ ⎠ . . . (5.10)
12 2
2
NV VN
⎛ ⎞′ = ⎜ ⎟
⎝ ⎠ . . . (5.11)
and 12 2
2
NI IN
⎛ ⎞′ = ⎜ ⎟
⎝ ⎠ . . . (5.12)
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AC MachinesSimilarly we can represent the equivalent circuit referred to secondary side as shown
below in Figure 5.4.
Figure 5.4 : Equivalent Circuit of a Transformer Referred to the Secondary Side
where 2
1
2i i
NGN
⎛ ⎞′ = ⎜ ⎟
⎝ ⎠G . . . (5.13)
2
1
2m m
NB BN
⎛ ⎞′ = ⎜ ⎟
⎝ ⎠ . . . (5.14)
2
21 1
1
NR RN
⎛ ⎞′ = ⎜ ⎟
⎝ ⎠ . . . (5.15)
2
21 1
1
NX XN
⎛ ⎞′ = ⎜ ⎟
⎝ ⎠ . . . (5.16)
21 1
1
NV VN
⎛ ⎞′ = ⎜ ⎟
⎝ ⎠ . . . (5.17)
and 11 1
2
NI IN
⎛ ⎞′ = ⎜ ⎟
⎝ ⎠ . . . (5.18)
SAQ 2 (a) Draw the equivalent circuit of a transformer referred to the secondary side.
(b) Find equivalent resistance and reactance of a transformer referred to primary.
5.2.3 Open Circuit Test and Short Circuit Test We must know the losses of a transformer to determine its efficiency. We shall discuss the two main losses in the transformer given below.
Copper Losses or Ohmic Losses (I2 R –loss)
These losses occur in the primary and secondary winding resistances. These losses vary as the square of the current (loading).
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Electrical
Core Loss or Iron Loss
Core loss comprises eddy current loss and hysteresis loss resulting from alternations of magnetic flux in the core.
Pe = Eddy current losses = Ke f2 B2 W/M3 . . . (5.19)
and Ph = Hysteresis losses = Kh f B1.6 W/M3 . . . (5.20)
where Ke and Kh are constants.
f → Supply frequency, Hz
B → Flux density, Wb/M2
The eddy current losses can be minimised by using the laminated core steel and the hysteresis losses can be minimised by adding silicon content to the iron core.
Open circuit test and short circuit tests are performed to determine the iron losses and copper losses respectively.
Open Circuit Test
This test is performed to find the core losses and shunt branch parameters Gi and BBm of the equivalent circuit of the transformer. This test is performed on the low voltage winding side while high voltage winding is kept open circuited. The connection diagram is shown in the Figure 5.5. Rated voltage is applied to the low voltage winding and meter readings are recorded.
Voltmeter reading = V1 Volts
Ammeter reading = I0 Amps
Power input = P0 Watts
Figure 5.5 : Connection Diagram for Open Circuit Test
The no load current I0 drawn by the transformer in very low usually 2-6% of the rated current. The winding resistances and reactances are of very low ohmic value. Therefore, ohmic losses can be neglected.
The equivalent (approximate) circuit is shown in Figure 5.6.
Figure 5.6 : Equivalent Circuit for Open Circuit
The watt meter reads the core losses (P0), watts :
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AC Machines . . . (5.21) 2
1 0iV G P=
02
1i
PGV
= . . . (5.22)
00
1
IY
V= . . . (5.23)
2 20m iB Y G= − . . . (5.24)
We should know the Gi and BBm are referred to the low voltage winding side. We can easily transform them to the HV winding side.
Short Circuit Test This test is performed to determine the copper losses and the series parameters (Req and Xeq) of the equivalent circuit of the transformer. This test is conducted on the HV winding side and LV winding in short circuited. The connection diagram is shown in Figure 5.7. A very low voltage (5-8% of the rated voltage) is sufficient to circulate the full load current under short circuit. The shunt branch parameters can be neglected in this case as the I0 is very small (0.1 to 0.5%) under the condition of very low voltage. While conducting the SC test, the voltage on HV winding side is gradually raised from zero to till the value transformer draws full load current. Record the meter readings.
Figure 5.7 : Connection Diagram for Short Circuit Test
Voltmeter reading = Vsc volts Ammeter reading = Isc Amps Power input = Psc Watts
Figure 5.8 : Equivalent Circuit for Short Circuit Test
The wattmeter reads the copper losses (Psc), watts
eqsc
sc
VZ
I= . . . (5.25)
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eq 2( )sc
sc
VRI
= . . . (5.26) Electrical
2 2eq eq eqX Z R= − . . . (5.27)
Req and Xeq are the ohmic values referred to the HV winding side. Thus short circuit test and open circuit test gives the parameters of the equivalent circuit of the transformer and also the efficiency and voltage regulation of the transformer.
Example 5.1 A 50 kVA, 2200/110 V, 50 Hz transformer has an HV winding resistance of 0.15 Ω and a leakage reactance of 0.45 Ω. The LV winding has a resistance of 0.006 Ω and a leakage reactance of 0.012 Ω. Find the equivalent winding resistance, reactance and impedance referred to the HV and LV sides.
Solution
1 1 1 0.15 0.45Z r j x j= + = +
2 2 2 0.006 0.012Z r j x j= + = +
1 2( )Z HV Z Z ′= +
2
22200 (0.006 0.012) (2.4 4.8)110
Z j⎛ ⎞′ j= + = +⎜ ⎟⎝ ⎠
Ω
( ) (0.15 0.45) (2.4 4.8) (2.55 5.25)Z HV j j j= + + + = + Ω
Similarly 1 2( )Z LV Z Z′= +
2110 (0.15 0.45) (0.006 0.012)
2200j j⎛ ⎞= + + +⎜ ⎟
⎝ ⎠Ω
(0.006375 0.01312)j= + Ω
SAQ 3 (a) Why do we neglect the copper losses in an open circuit test? (b) A 50 KVA, 2200/110 V transformer when tested gave the following
results : Open circuit test (LV side) 400 W, 9 A, 110 V Short circuit test (HV side) 808 W, 20.5 A, 88 V
Compute all the parameters of the equivalent circuit referred to the HV and LV sides of the transformer.
5.3 INDUCTION MOTORS
The first crude type of poly phase induction motor was presented at the Frankfurt exhibition by Nikola Tesla in 1891. Since then significant improvements have been made in the design and quality of the induction motors and a large variety of single phase and three phase induction motors have been developed. In this section, we shall discuss the constructional details, theories and operating characteristics of poly phase induction motors.
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AC Machines5.3.1 Constructional Details
An induction motor has two main parts, i.e. stator and rotor. An induction motor has the same physical stator as a synchronous machine. The stator and rotor both are cylindrical. The magnetic flux is produced by the stationary winding housed in the outer part, the stator, to which the supply is connected whereas the inner part, the rotor, which is also called armature, carries rotor winding in which the current is induced.
Stator
The stator core is made up of laminations which may or may not be insulated by varnish. Thicker laminations are used in small motors. A partially wound stator of induction motor is shown in Figure 5.9.
Figure 5.9 : A Partially Wound Stator of Induction Motor
The stator of poly phase induction motor carries the poly phase AC windings. Stator winding is completely insulated as per the voltage ratings. We shall generally use the double layer winding owing to the ease of manufacture, assembly and repair.
Rotor
There are two different types of induction motor rotors which can be placed inside the stator. One is called squirrel cage rotor or simply a cage rotor, while the outer is called a wound rotor. Figure 5.10 shows squirrel-cage induction motor rotor and Figure 5.11 shows wound rotor of poly phase induction motor.
Figure 5.10 : Sketch of Squirrel-cage Rotor
A squirrel-cage induction motor rotor consists of a series of conducting bars laid into slots carved in the face of the rotor and shorted at either end by large shorting rings. The squirrel cage rotor are simple, more rugged and cheaper than wound rotors.
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Electrical
Figure 5.11 : Wound Rotor of Poly Phase Induction Motor
A wound rotor has a complete set of three-phase windings that are the mirror images of the windings on the stator. The three-phases of the rotor windings are usually, Y-connected, and the ends of the three rotor wires are ties to slip rings on the rotor’s shaft. Wound rotor induction motors therefore have their rotor currents accessible at the stator brushes, where they can be examined and where extra resistance can be inserted into the rotor circuit.
Wound rotor induction motors are more expensive than squirrel-cage induction motors and they require more maintenance because of the wear associated with their brushes and slip rings.
Therefore on the basis of rotors construction, the induction motors are classified as :
(a) Squirrel-cage induction motor.
(b) Phase wound or slip ring induction motor.
5.3.2 Rotating Magnetic Field
We have discussed the general construction of poly phase induction motors. Now we shall discuss the concept of rotating magnetic field in poly phase induction machines.
Three identical and individual windings, displaced on the stator by 120o and carrying current also displaced 120o in phase or time are required to establish a constant rotating magnetic field. This magnetic field rotates at synchronous speed (Ns = 120 f/P). Figure 5.12 shows the stator of a three-phase induction machine. For simplicity, we shall assume the winding concentrated in six (6) slots.
Figure 5.12 : A Generalised Configuration of a Two Pole Three-phase Stator Winding
These concentrated full pitch coils represents the distributed windings producing sinusoidal mmf waves. R, Y and B represent the start end and R′, Y′ B′ represent the finish
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AC Machinesend of three phases R, Y and B respectively. We know that the windings of the individual
phases are displaced by 120 electrical degree in the space around the air gap therefore the mmfs of these windings will also be displaced by 120o in space.
FR, FY and FB represent the mmfs of the three phases, when currents in them are at their positive peak values. Each phase current is alternating sinusoidally with time. As the currents alternate, the mmfs of the three phases will vary in magnitude and change direction so as to follow the change in current.
B
The air gap mmf at any angle, θ, is due to contribution by all the three phases. Let the angle θ be measured from the axis of phase R. The resultant mmf at angle θ is
. . . (5.28) o o(peak) (peak) (peak)[ cos cos ( 120 ) cos ( 120 )]R Y BF F F Fθ = θ + θ − + θ +
Starting from the instant when current in phase R is at its maximum value, we have
FR (Peak) = FR (max) cos ωt . . . (5.29)
Fy (Peak) = Fy (max) cos (ωt – 120o) . . . (5.30)
FB (Peak) = FB (max) cos (ωt + 120o) . . . (5.31)
For balanced currents,
FR (max) = Fy (max) = FB (max) = Fmax
Substituting the values from Eqs. (5.29), (5.30) and (5.31) into Eq. (5.29), we have
o omax max( , ) cos cos cos ( 120 ) cos ( 120 )F t F t F tθ = ω θ + ω − θ −
. . . (5.32) max cos ( 12) cos ( 120)F t+ ω + θ +
Solving the Eq. (5.27), we have
max( , ) = 1.5 cos ( )F t Fθ tθ − ω . . . (5.33)
The resultant mmf wave given by Eq. (5.33) has a constant magnitude. The term (ωt) gives rotation of mmf around the air gap at a constant angular velocity ω. The angular velocity of the wave is ω = 2 π f electrical radians per second.
5.3.3 Principle of Operation of a Three-phase Induction Motor When the stator windings of a three-phase induction motor are connected to a three-phase balanced supply, the three-phase balanced currents develop a rotating flux wave as we have discussed in the previous section. The rotating flux wave cuts the rotor conductors and induces voltage in the rotor conductors. As the rotor conductors are short-circuited these induced EMFs develop current in the rotor. These currents and the flux wave interact to produce torque in the rotor. As per the lenz’s law, the developed torque makes the rotor to rotate in the direction of the flux wave so as to reduce the relative speed between the flux wave and the rotor conductors and thereby reducing the cutting of flux lines by rotor conductors.
The rotating flux wave produced by the stator rotates at a speed of Ns (RPM) with respect to stator, where synchronous speed Ns is given by
120s
fNP
= . . . (5.34)
where f = Stator voltage frequency, Hz, and
P = Number of poles in the stator windings.
The rotor speed Nr of an induction motor is always slightly less than the synchronous speed Ns.
The relative speed between the rotating flux and the rotor is usually represented as a fraction of the synchronous speed which is called slip.
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Slip s r
s
N NS
N−
= . . . (5.35) Electrical
The speed of induction motor rotor is
(1 )r sN N S= − . . . (5.36)
The rotor induced emf will have the frequency
2f S f= . . . (5.37)
Since the rotor frequency is directly proportional to the slip, f2 is often called the slip frequency.
Example 5.2
Two three-phase induction motors when connected across a 400 V, 50 Hz supply are running at 1440 and 940 RPM respectively. Determine which of the two motors is running at higher slip.
Solution
We know that f = 50 Hz and we can determine the synchronous speed for
120 502, 3000 RPM2sP N ×
= = =
120 504, 1500 RPM4sP N ×
= = =
120 506, 1000 RPM6sP N ×
= = =
An induction motor runs at a speed slightly less than the synchronous speed.
Numbers of poles of the motor running at 1440 rpm should be 4 and Ns = 1500 rpm.
% Slip of the motor (A),
1500 1440100 1000 = 4%1500
− −= × = ×s r
As
N Ns
N
Numbers of poles of the motor running at 940 rpm should be six and Ns = 1000 rpm.
The slip of the motor (B),
1000 940 = 6%1000−
=Bs
Thus the slip of the motor running at 940 rpm is higher than the slip of the motor running at 1440 rpm.
SAQ 3 A three-phase, six pole induction motor is supplied from a 50 Hz, 400 V supply. Calculate (a) the synchronous speed, and (b) the speed of the rotor when the slip is 3%.
5.3.4 Phasor Diagram of an Induction Motor Induction motor is analogous to the transformer. For the stator winding
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AC Machines
1
1
. . . (5.38) 1 1 1 1( )V E I R j X= − + +
where V1 = Supply voltage per phase,
I1 = Stator current per phase,
R1 = Stator resistance per phase,
X1 = Stator leakage reactance per phase, and
E1 = Voltage induced per phase in the stator.
and also,
1 14.44E f N kw= φ . . . (5.39)
where φ = Mutual flux per pole,
f = Stator frequency,
N1 = Stator winding turns per phase, and
kw1 = Winding factor of stator.
We know that rotor winding is short circuited at the ends
2 2 2 20 ( 2 )s sV E I R j X= = − + . . . (5.40)
where I2 = Rotor current per phase,
R2 = Rotor resistance per phase, and
X2s = Leakage reactance of rotor at slip frequency per phase.
2 22s 2X f L= π ;
2f S f=
2 22s 2X S f L= π
2 2sX S X=
where X2 = Leakage reactance of rotor as stand fill per phase, and
E2s = Induced emf per phase in the rotor of induction motor.
2 2 24.44s 2E f N Kw= φ . . . (5.41)
= S E2
where E2 = Induced emf in rotor per phase at stand still,
N2 = Rotor winding turns per phase, and
Kw2 = Winding factor of rotor.
From the Eqs. (5.39) and (5.41), we have
1 1 1 1
2 2 2 2
E N Kw N aE N Kw N
′= =
′=
2N ′
1 |
. . . (5.42)
where, are the effective number of stator and rotor turns respectively and “a” is called the effective turns ratio of an induction motor.
1 andN ′
. . . (5.43) 1| | |V E≈
since the value of R1 and X1 are small.
The exciting current in the induction motor is given by
0 i mI I I= + . . . (5.44)
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Electrical
where mI is the per phase value of the magnetising component of the exciting current and iI is the per phase value of the loss component of the exciting current.
The no load current in induction motor is more than that of transformer because of the air gap between the stator and rotor winding.
From the voltage and current equations discussed above, we can develop the phasor diagram.
The rotor power factor angle at any slip, S is
1 22
2tan S X
R− ⎛ ⎞
θ = ⎜ ⎟⎝ ⎠
. . . (5.45)
Figure 5.13 : Phasor Diagram of an Induction Motor for a > 1 and S < 1
where 22 2
1
1NI IN a 2I′
′ = =′
. . . (5.46)
and 1 0I I I2′= + . . . (5.47)
5.3.5 Equivalent Circuit
We know that 1 2
2 2
1andE IaE I a
′= = .
The impedance of the rotor circuit
2 2 2Z R j S X= + . . . (5.48)
147
AC Machinesand 2 2 2Z R j X= + (at standstill)
Figure 5.14 : Rotor Circuit at Frequency F2 = Sf
22
2 2( )S EI
R j S X=
+ . . . (5.49)
2 22 2 2Z a R j S a X′ = + . . . (5.50)
2 2 2Z R j S X′ ′= + ′
which gives
1 12
22 22
S E EI RR j S X j XS
′ = = ′′ ′+ ′+ . . . (5.51)
The Eq. (5.51) refers the rotor circuit to the stator frequency f. In referring the rotor circuit to the stator frequency the reactance becomes constant ( 2X ′ ) and the resistance
becomes variable 2RS′⎛
⎜⎝ ⎠
⎞⎟ . The equivalent circuit of an induction motor is shown in
Figure 5.15.
Figure 5.15 : Equivalent Circuit of an Induction Motor Referred to Stator Side
If 2R′ is separated from 2′RS
to represent the rotor copper loss as a separate quantity, the
circuit model can be redrawn as Figure 5.16 where 21 1⎛′ ⎞−⎜⎝ ⎠
Rs ⎟ represents the mechanical
output in electrical form.
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Electrical
Figure 5.16 : Equivalent Circuit of an Induction Motor
5.3.6 Air Gap Power, Torque and Output Power
As shown in Figure 5.17, the power crossing the terminals 1-2 is the electrical power input per phase minus the stator copper loss and iron loss. This is the power transferred from the stator to the rotor via the air gap magnetic field. The PG is the three-phase value and known as air gap power.
It is obvious from the equivalent circuit that
2 223G
RP IS′⎛′= ⎜
⎝ ⎠⎞⎟ . . . (5.52)
Power across air gap Rotor copper lossSlip
=
Figure 5.17 : Equivalent Circuit of Induction Motor
Thus mechanical power output (Gross)
22 23m GP P I R′ ′= −
22 2
13 1 (1 GI R S PS
⎛ ⎞′ ′= − = −⎜ ⎟⎝ ⎠
) . . . (5.53)
Rotor speed is
(1 ) rad (mech)/secr sSω = − ω
The electromagnetic torque developed is then given by
(1 ) (1 )s e mS T P S PG− ω = = −
or
1 223
NmGe
s s
RIP ST
′⎛ ⎞′ ⎜ ⎟⎝ ⎠= =
ω ω . . . (5.54)
149
AC MachinesThe net mechanical power output and torque are obtained by subtracting losses-windage,
friction and stray load loss.
Example 5.3
A 6-pole, 50 Hz, 3-phase induction motor running on full load develops a useful torque of 180 Nm when the rotor emf frequency is 2 Hz. Calculate the shaft power output. If the mechanical torque lost in friction and that for core-loss is 12 Nm, calculate
(a) the input to the motor, and
(b) efficiency.
The total stator loss is 900 W.
Solution
2 2f S f= =
2 0.0450
S = =
120 50 1000RPM6sN ×
= =
(1 ) (1 0.04) 1000 960 RPMr sN S N= − = − =
2 960 100.53 rad/s60r
π ×ω = =
Shaft power output 180 100.53 18.095 kW= × =
Mechanical power developed (180 12) 100.53 19.301 kW= + × =
1Rotor copper loss 1mPS
⎛ ⎞= −⎜ ⎟⎝ ⎠
Rotor copper loss 0.0419.301 0.804 kW1 1 0.04m
SPS
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠
(a) Input to the motor = 19.301 + 0.804 + 0.9 = 21.005 kW
(b) Efficiency of motor 18.095 100 86.14%21.005
= × =
SAQ 4 An 8 pole, 3-phase, 50 Hz induction motor runs at a speed of 710 RPM with an input power of 35 kW. The stator copper loss at this operating condition is 1200 W while the rotational losses are 600 W. Find
(a) Rotor copper loss.
(b) Gross torque developed.
(c) Gross mechanical power developed.
5.3.7 Torque-Slip (T-S) Characteristics For the better expression of the torque make thevenin’s equivalent circuit for the left side of the terminal 1-2.
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Electrical
Figure 5.18 : Approximate Equivalent Circuit
The thevenin impedance.
1 1( ) | |T m T TZ R j X j X R j X= + = +
Thevenin voltage 11 1
mT
m
j XV V
R j X j X⎡ ⎤
= ⎢ ⎥+ +⎣ ⎦
Figure 5.19 : Thevenin’s Equivalent
2 222
2( )
T
T T
VIRR XS
′ =′⎛ ⎞ X ′+ + +⎜ ⎟
⎝ ⎠
. . . (5.55)
2 22
3
s
RT IS′⎛ ⎞′= × ⎜ ⎟ω ⎝ ⎠
. . . (5.56)
Substitute the value of 2I ′ in Eq. (5.56)
2 2
222
2
3 .( )
T
sT T
RVST
RR XS
′⎛ ⎞⎜ ⎟⎝ ⎠=
ω ′⎛ ⎞ X ′+ + +⎜ ⎟⎝ ⎠
. . . (5.57)
Eq. (5.57) expresses the torque as a function of voltage and slip. It is seen that for a fixed value of slip, torque is proportional to the square of voltage. Figure 5.20 shows the torque slip characteristics of an induction motor for the rated voltage.
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AC Machines
Figure 5.20 : Torque-Slip Characteristics
The torque is zero when slip is zero; (Nr = Ns). We have already discussed that induction motor torque is zero at synchronous speed. At slip Sm, T; the motor gives the maximum torque that is also called break down torque. When the rotor is stand still (S = 1), the torque corresponds to starting torque Ts. In a normally designed motor Ts is much less than Tmax (Break down torque). The torque slip characteristics from no-load to somewhat beyond full-load is almost linear. The condition for maximum torque
2, 2 2
2( )m T
T T
RSR X X
′=
′+ + 2 . . . (5.58)
where SM, T is the value of slip at which maximum torque occurs, and the maximum torque is then
2
max 2 22
(0.5 )3 .( )
T
s T T T
VTR R X X
=ω ′+ + +
. . . (5.59)
From the Eq. (5.59), we came to the conclusion that the maximum value of the torque does not depend upon rotor resistance 2( )R′ .
The slip at which maximum torque occurs indirectly proportional to the rotor resistance 2( )R′ . At S = 1, we have the starting torque which increases by adding resistance in the rotor circuit.
Figure 5.21 : The Effect of Rotor Resistance on Torque-Slip Characteristics
5.3.8 Starting of Induction Motors
At the time of starting S = 1, the starting current can be as large as five to six times the full-load current. With such a large starting current, the motor must accelerate and reach normal speed quickly otherwise overheating may damage the motor. Therefore, it is only small size motors that can be started direct on-line. Either a full voltage or reduced voltage is applied across the stator terminals to start the induction motors. The reduced voltage limits the starting current but at the very same time it reduces the starting torque.
Squirrel Cage Motor Starting
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Electrical
We know that there are several methods of starting the squirrel cage induction motors, i.e.
(a) Direct on-line starting
(b) Stator resistor starting
(c) Auto-transformer starting
(d) Star-delta starting
Direct On-line Starting
In this method, the three-phase stator is directly switch “ON” to the three-phase supply mains. The motor takes starting current of 5 to 6 times its full load current depending upon the design of motor. Such a starting causes voltage drops in the power supply lines feeding the induction motor.
Stator Resistor Starting
In this method, we connect a resistor or a reactor in between motor terminals and the supply. As we start the motor, some voltage drop occurs across the starting resistor or reactor and only a fraction of the supply voltage appears across the stator terminals. Thus, it reduces the starting current. As the motor speeds up the resistor/reactor is cut out in steps and finally short circuited.
Auto-transformer Starting
A fraction KV1, of the supply voltage V1 is applied to the stator at the time of starting by means of auto-transformer.
Figure 5.22 : Auto-transformer Starting (k < 1)
This limits the starting current to k Is. As the motor attains its full speed, the full voltage is set through the auto-transformer.
The starting current and starting torque are reduced to k2 times their corresponding values with direct-on-line starting.
Star-Delta Starting
This method is used for delta operated motors. The stator phases are first connected in star by a special switch (triple pole double throw switch). As the steady state speed is reached, the stator is connected in delta by changing the position of the same switch. The star-delta starter reduces the starting torque to one-third of that produced by direct switching in delta. Star-Delta
153
AC Machinesstarter is cheap as compared to auto-transformer starter and is therefore used
extensively.
Starting of Wound-rotor Motors
The simplest and cheapest method of starting wound-rotor induction motors is by inserting the external resistance in the rotor circuit with full line voltage. The addition of resistance in the rotor circuit of a wound-rotor induction motor decreases the starting current, increases the starting torque and improves the starting power factor. The external rotor resistance is arranged in steps which are gradually cut out during starting.
SAQ 5
(a) Why do we require the starter for the starting of induction motors.
(b) What are the different methods of starting? Explain at least one.
5.3.9 Speed Control of Three-phase Induction Motors Until the advent of modern solid-state drives induction motors in general were not good machines for applications requiring considerable speed control. The normal operating range of a typical induction motor is confined to less than 5% slip and the speed variation over that range is more or less directly proportional to the load on the shaft of the motor. Even if the slip could be made larger, the efficiency of the motor would become very poor.
There are actually two techniques by which the speed of an induction motor can be controlled. One is to vary the synchronous speed, which is the speed of the stator and rotor magnetic fields, since the rotor speed always remains near Ns. The other technique is to vary the slip of the motor for a given load.
The synchronous speed of the machine is
120s
fNP
=
The synchronous speed can be varied by
(a) changing the electrical supply frequency.
(b) Changing the number of poles on the machine.
Slip control may be accomplished by varying either the rotor resistance or the terminal voltage of the motor.
There are two major approaches to change the number of poles in an induction motor.
(a) The method of consequent poles.
(b) Multiple stator windings.
The first method is an old method for speed control developed in 1897. The number of poles in the stator windings of an induction motor can easily be changed by a factor 2 : 1 with only simple changes in coil connections. The rotor in such motor is of the squirrel cage design. The traditional approach to overcome the limitations of first approach was to employ multiple stator windings with different number of poles and to energise only one set at a time.
By using variable frequency control, it is possible to adjust the speed of the motor either above or below the base speed. It can control the speed of an induction motor over a range from as little as 5 per cent of base speed up to about twice base speeds.
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Electrical
In wound-rotor induction motors, it is possible to change the shape of the torque-speed curve by inserting extra resistances into rotor circuit of the machine. However, inserting extra resistances into the rotor circuit of an induction motor seriously reduces the efficiency of the machine. Such a method of speed control is normally used only for short periods because of this efficiency problem.
5.3.10 Applications of Induction Motors squirrel cage induction motor is used for the loads require low starting torques and substantially constant speed. Squirrel cage motors with relatively low rotor resistance are used for fans, centrifugal pumps, etc. Cage motors with relatively high rotor resistance are used for crushers, compressors, etc.
A wound-rotor motors are used for hoists, cranes, elevators, compressors, etc.
SAQ 6 (a) How can we control the speed of induction motors?
(b) Give the name of different techniques to control the speed of induction motors.
5.4 SINGLE PHASE AND SPECIAL PURPOSE MOTORS
In this section, we shall discuss some special types of motors, i.e. universal motor, permanent magnet DC motor, AC and DC servomotors. However, most homes and small businesses do not have three-phase power available. For such locations, all motors must run from single-phase power sources.
5.4.1 Single Phase Induction Motors A single phase induction motor has a distributed winding on the stator and normal squirrel-cage rotor as shown in Figure 5.23. When the rotor is at rest, there is no torque. When fixed stator coil carries alternating current, the MMF wave produced is stationary in space but pulsates in magnitude and varies sinusoidally with time. The currents induced in rotor are in such a direction as to oppose the stator MMF. The rotor currents in the left half are directed opposite to those in the other half. The net torque on the rotor due to interaction of these currents with stator MMF in the vertical axis is zero which results zero torque.
There are two methods of analysing this motor, vis., cross field theory and rotating field theory. The treatment of single-phase induction motor is based on the revolving field theory. The revolving field theory gives a very easy physical understanding of the topics. Single-phase induction motors suffer from a severe handicap. Since there is only one phase on the stator winding, the magnetic field in a single-phase motor does not rotate. Instead, it pulses, getting first larger and then smaller, but always remaining in the same direction. Because there is no rotating stator magnetic field, a single-phase induction motor has no starting torque. However, once the rotor begins to turn, and induced torque will be produced in it.
155
AC Machines
Figure 5.23 : Elementary Single-phase Induction Motor
As seen from the axis of the winding, the MMF at any angle, θ, is
. . . (5.60) cosPF F= θ
where FP = Peak MMF, and
θ = Angle measured from winding axis.
We know that
cosP MF F t= ω . . . (5.61)
where FM = Maximum value of MMF.
From Eqs. (5.60) and (5.61), we have
1 1cos ( ) cos ( )2 2M MF F t F= θ − ω + θ + tω . . . (5.62)
where 1 cos ( )2 MF tθ − ω is the forward rotating field, Ffd, and
1 cos ( )2 MF tθ + ω is the backward rotating field, Fbd.
Figure 5.24 : Two Components of a Pulsating Field
If the rotor rotates at the speed Nr, in the direction of Ffd, the slip of rotor with respect to the Ffd is then
s rfd
s
N NS S
N−
= = . . . (5.63)
while the rotor slip with respect to the Fbd is
( )
(2 )s rbd
s
N NS S
N− −
= = − . . . (5.64)
Thus the rotor slips with respect to the backward and forward rotating fields are different.
156
The torque-speed characteristics of a single-phase induction motor is shown in Figure 5.25.
Electrical
Figure 5.25 : Torque-speed Characteristics of a Single Winding Single-phase Induction Motor
When Nr = 0; S = 1, the two rotating fields have the same strength and produce equal and opposite torques resulting in net starting torque of zero value. If, however, the rotor is made to run at speed Nr in the direction of the forward field, the two slips are how S and (2 – S). The result of weakening of one field and simultaneously strengthening of the other leads to a torque-speed characteristics like that of a three-phase induction motor in the speed region close to Ns.
Single-phase induction motors may be classified in the following way
(a) Resistance split-phase motor
(b) Capacitor split-phase motor
(c) Capacitor-start motor
(d) Two value capacitor motor.
5.4.2 The Universal Motor
The induced torque of a DC motor is given by
aT k I= φ
If we reverse the polarity of the voltage applied to a shunt or series motor, the resultant induced torque continues in the same direction because both the direction of the field flux and the direction of the armature current is reversed. Therefore, it is possible to achieve a pulsating but unidirectional torque from a DC motor connected to an AC power supply. Practically it is possible for the DC series motor, since the armature current (Ia) and the field current (If) must reverse at exactly the same time. In order for a series DC motor to function effectively on AC, its field poles and stator frame must be completely laminated, otherwise their core losses would be enormous. When the poles and stator are laminated, this motor is often called a universal motor, since it can run from either AC or DC supply. An equivalent circuit of a universal motor is shown in Figure 5.26.
157
AC Machines
Figure 5.26 : Equivalent Circuit of a Universal Motor
The armature and field windings have quite a large reactance at 50 Hz. Therefore, Ea becomes smaller for AC supply as compared to DC supply. The motor is slower for a given armature current and induced torque on AC than it would be on DC. A typical torque-speed characteristics of a universal motor is shown in Figure 5.27.
Figure 5.27 : Speed-torque Characteristics of a Universal Motor
Applications
It is not suitable for constant speed applications. It is compact and gives more torque per ampere than any other single-phase motor. It is used where light weight and high torque are important. Some applications of this motor are vacuum cleaners, drills, portable tools and kitchen appliances.
5.4.3 Permanent Magnet DC Motor (PMDC)
A permanent magnet DC motor is a DC motor whose poles are made of permanent magnets. Since these motors do not require an external field circuit, they do not have the field circuit copper losses associated with DC shunt motors. They are compact than corresponding DC shunt motors. These motors are common in smaller fractional and sub-fractional horse power sizes, where the expense and space of a separate field circuit can not be justified. Permanent magnets can not establish as high a flux density as an externally supplied shunt field, so these motors will have a lower induced torque.
In a PMDC machine, the pole flux is just the residual flux in the permanent magnets. If the armature current becomes very large, there is some risk that the armature mmf may demagnetize the poles, permanently reducing or re-orienting the residual flux in them.
A permanent magnet DC motor is basically the same machine as a DC shunt motor except that the flux of a PMDC motor is fixed. A very general diagram of a PMDC motor is shown in Figure 5.28.
Since it uses the permanent magnet poles, it is not possible to control the speed of a PMDC motor by varying the field current or flux.
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Electrical
Figure 5.28 : A Very General Diagram of PMDC Motor
5.4.4 AC Servomotor AC motor are preferred for low power application as they are light weight and rugged. Two phase induction motors are used in feedback control system.
Figure 5.29 : AC Servomotor
The motor consists of a stator with two distributed windings displaced 90 electrical degree apart. Under normal operating conditions, a fixed voltage from a constant voltage source is applied to one phase, the other phase is energized by a voltage of variable magnitude and polarity. This voltage is at 90o out of phase with respect to the voltage of fixed phase. The control phase voltage is usually supplied from the servoamplifier. The direction of rotation of the motor reverses if control phase signal changes sign. The rotor construction is usually of squirrel cage or drag-cup type with no electrical access. The torque-speed characteristics of a servomotor (two-phase induction motor) is shown in Figure 5.30.
159
AC Machines
Figure 5.30 : Speed Torque Characteristics of AC Servomotor
5.4.5 DC Servomotor
The DC motors are experience because of brushes and commutators. These motors have relatively lower torque to volume, and torque-to-inertia ratios. However, the characteristics of DC motors are quite linear and these motors are easier to control.
The sketch of the basic components of DC servomotor is shown in Figure 5.31. DC servomotor is essentially an ordinary DC motor with a few difference in the construction. In servo application, DC motor is required to produce rapid acceleration from standstill. The requirement of DC servomotor are low inertia and high starting torque. Low inertia can be obtained by reducing the armature diameter and increasing the length of the armature core.
Two different modes in which DC motor can be operated are :
(a) Field control mode.
(b) Armature control mode.
Figure 5.31 : DC Servomotor
In a permanent magnet motor, the flux (φ) is constant, the torque exerted on the motor rotor can therefore be controlled by varying the armature current. If the direction of the armature current is reversed, the direction of the torque is reversed.
These motors are used in many low power control applications.
SAQ 7 (a) What is the universal motor?
(b) Where do we use the PMDC motor?
(c) How does a DC servomotor differ from an ordinary DC machine?
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Electrical 5.5 SUMMARY
In this unit, we have learnt the basic concepts of the transformer, induction motors and special type fractional kilowatt (kW) machines. In the section of transformer you have studied the basic construction, operation principle and testing of the transformers. You have also learnt about the induction motor construction and working principle. Further in the section of induction motors we have focused on the starting and speed control of induction motors, applications of the induction motors. In the section of single-phase and special purpose motors you have learnt about universal motors, servomotors and permanent magnet DC motors.
5.6 ANSWERS TO SAQs
SAQ 1 (a) Transformers are used most widely in electric power systems to change the
voltage levels. We need the transformers for (i) Changing voltage and current levels. (ii) Matching source and load impedances for maximum power transfer in
electronic and control circuits. (iii) Electrical isolation (isolating one circuit from another).
(b) Refer Section 5.2.1 (Working Principle). (c) Step up transformer is used to step up the voltage. The secondary voltage
(V2) is greater than the primary voltage (V1). The step down transformer is used to step down the voltage. The secondary voltage is less than the primary voltage.
N2 > N1, step up transformer N2 < N1, step down transformer
(d) The main components of the transformer are primary and secondary windings and the magnetic core. For large rating transformer this whole assembly is kept inside a tank. A transformer oil is filled in the tank for insulation and cooling.
SAQ 2 (a) Refer Figures (5.2) and (5.4). (b) Equivalent resistance reactance of a transformer referred to primary side are
21
eq,1 1 22
NR R RN
⎛ ⎞= + ⎜ ⎟
⎝ ⎠
21
eq,1 1 22
NX X XN
⎛ ⎞= + ⎜ ⎟
⎝ ⎠
SAQ 3 (a) The no load current (I0) drawn by the transformer is very low 2-6% of the
rated current. The winding resistances and reactances are of very low ohmic value. Therefore copper losses can be neglected in an open circuit test.
(b) OC test (LV side)
09 0.0818
110y = =
i 2400 0.033
(110)G = =
161
AC Machines2 2
0 0.0748m iB y G= − =
SC test (HV side)
88 4.2920.5
Z = = Ω
2808 1.922
(20.5)R = = Ω
2 2(4.29) (1.922) 3.835X = − = Ω
Now 2
5110( ) (0.033) 8.25 102200iG HV −⎛ ⎞= = ×⎜ ⎟
⎝ ⎠
2
4110( ) (0.0748) 1.87 102200mB HV −⎛ ⎞= × = ×⎜ ⎟
⎝ ⎠
and 2
3110( ) (1.922) 4.8 102200
R LV −⎛ ⎞= = ×⎜ ⎟⎝ ⎠
Ω
2
3110( ) (3.835) 9.5 102200
X LV −⎛ ⎞= = ×⎜ ⎟⎝ ⎠
Ω
SAQ 4
120 120 50 1000 RPM6s
fNP
×= = =
Percentage slip, ( )
100s r
s
N NS
N−
= ×
10003 11000
rN−= × 00
970 RPMrN =
SAQ 5
120 120 50 750 RPM8s
fNP
×= = =
2 750 78.54 rad/s60s
π ×ω = =
710 RPMrN =
Slip 750 710 0.053750−
= =
74.38 rad/secrω =
Electrical power input = 35 kW
Stator copper loss = 1.2 kW
35 1.2 33.8 kWGP = − =
(a) Rotor copper loss 0.053 33.8 1.79 kW= = × =Gs P
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Electrical (b) Gross torque
333.8 10 430.4 Nm78.54×
= =
(c) Mechanical power developed (gross) (1 )= − Gs P
(1 0.053) 33.8 32 kW= − × =
SAQ 6
(a) If we switch on the large rating machines directly to the power supply, they draw heavy current at the time of starting (five to six times the full load current). This heavy current causes the over heating of the machine and also voltage drops in the supply line. Therefore, to limit the starting current, a starter is required.
(b) Refer Section 5.3.8.
SAQ 7
(a) There are actually two techniques by which the speed of an induction motor
can be controlled. One is to vary the synchronous speed 120s
fNP
⎛ ⎞=⎜ ⎟⎝ ⎠
. The
other is to vary the slip of the motor for a given load.
(i) Changing the electrical supply frequency.
(ii) Changing the number of poles on the machine.
(b) The different techniques are
(i) The method of consequent poles.
(ii) Multiple stator winding.
(iii) Variable frequency supply.
(iv) Inserting the extra resistance into the rotor circuit.
SAQ 8
(a) Universal motor is a DC series motor with some modifications. Its field poles and stator frame are completely laminated. It can run either on AC or DC.
(b) Refer Section 5.4.2.
(c) A DC servomotor is essentially an ordinary DC machine with some slight modifications to give high starting and low inertia. Low inertia can be obtained by reducing the armature torque diameter and increasing the length of the armature core.
FURTHER READINGS Theraja and Theraja, A Text Book of Electrical Technology, Volumes I and II, S. Chand and Company.
Sahadev, Fundamentals of Electrical and Electronics Engineering, Dhanpat Rai and Company, New Delhi.
Chakrabarti A., Circuit Theory, Dhanpat Rai and Company, New Delhi.
163
AC MachinesHuges E. Pearson, Electrical Technology, Longmans.
Kothari D. P., Nagrath, I. J., Electric Machines, Third Edition, TMH, New Delhi.
Bhag S. Guru, Hiziroglu, Electric Machinery and Transformers, Third Edition, Oxford University Press, New York.
Kothari D. P. and Nagrath I. J. (2002), Basic Electrical Engineering, 2nd Edition, TMH, New Delhi.
Kothari D. P. and Nagrath I. J. (1998), Theory and Problems of Basic Electrical Engineering, Prentice Hall of India, New Delhi.
Kothari D. P. and Nagrath I. J. (2006), Electric Machines, Sigma Series, TMH, New Delhi.
ELECTRICAL This block consists of five units. In Unit 1 entitled ‘Properties of Conductors and Insulators’, you will be introduced with classification of material as conductor, insulator and semiconductor. You will be explained about some important properties of material such as resistance, inductance and capacitance.
164
In Unit 2 entitled ‘Electromagnetism’, our main focus will be on magnetic circuit, calculation of BH curve, permeability, hysteresis, concepts of self and mutual inductance and electromagnet.
Electrical
In Unit 3 entitled ‘DC Circuits”, would help you acquire a proper understanding of the basic concepts of the Kirchhoff’s law, Nodal and Mesh analysis, source transformation, Network theorems. Here you will apply these concepts on DC circuits to analyse them.
As most of the applications of electrical use electrical circuits with alternating quantities, here in Unit 4 entitled ‘AC Circuits’, you will be introduced to alternating quantities, average and RMS values of various waveforms, phasor representation RLC circuits, resonance. You will also learn about the concept of active power, reactive power and power factor, analysis of balanced three phase circuits.
Various AC machines play a major role in the today’s electrical world. In Unit 5 entitled ‘AC Machines’, you will learn the working principle, operational characteristics and applications of transformer; single and three phase induction motor; single phase motors – universal motor, permanent magnet DC motor, servo motor and their applications.
We hope that the course material presented here would help you to understand the principles of electrical science.
We wish you all the very best!
PRINCIPLE OF ELECTRICAL AND ELECTRONICS Basic knowledge of Electrical and Electronics is important to any engineering education. Applications of electrical and electronics have become important and integral part of our
165
AC Machinesdaily life. Therefore, knowledge of basic principles of them will definitely help us to
understand the technicality of electrical and electronics appliances.
The course contents are divided into two blocks covering principles of electrical and electronics.
We begin by Block 1 entitled Electrical, here in first part of this block various properties and characteristics of materials will be explained.
These materials are being classified as conductor, insulator and semiconductor. In this block the main emphasis will be on conductors and dielectrics or insulators. Some electrical properties like resistance, inductance and capacitance of these materials will also be explained.
In next units of this block you will acquire the knowledge of magnetic circuits under the unit of “Electromagnetism”. Kirchhoff’s law, Nodal and Mesh analysis, Theorems and Source transformation will be explained in Unit 3 : DC circuits. Introduction to alternating quantities, analysis of balanced three phase circuits and circuits containing resistance, inductance and capacitance will be analysed in the unit of “AC circuits”.
Electrical machines are largely used in domestic and industrial applications. In last unit of AC machines you will learn about different kinds of machines with their working principles and various operating characteristics.
In today’s modern society electronics has greatest impact. Block 2 : Electronics of this course contains four units. The first part of this block is devoted to semiconductor devices. The devices types covered include diodes, transistors, FET’s, power devices, SCR’s, MOSFET, IGBT’s, etc.
In next part of this block basic electronics circuits will be covered at length. This unit contains lot of circuits, practical and theoretical information that would be of particular interest to learners, practicing engineers and hobbyists. The last part of this block will introduce you with various aspects of microprocessors and recent trends in microprocessors. Digital electronics form the subject matter of next unit.
Further, a large number of solved example and self assessment question (SAQs) have been included to assist the students in attaining complete grasp of the subject.
