Electrical machines 2 AC Machines

ELECTRICAL MACHINES – II

(AC MACHINES)Presented by

C.GOKULAP/EEE

Velalar College of Engg & Tech,ErodeEMAIL: [email protected]

SyllabusEE6502 Electrical Machines -II

BOOKS Reference

LOCAL AUTHORS: For THEORY use this books1.Electrical Machines-II by “Gnanavadivel” – Anuradha Publication2. Electrical Machines-II by “Godse” – Technical Publication

For Problems:

Electric Machines by Nagrath & Kothari Refer Solved Problems Electric Machinery by A.E.Fitgerald Refer Solved Problems

Important Website Reference

Electrical Machines-II by S. B. Sivasubramaniyan -MSEC, Chennai

http://yourelectrichome.blogspot.in/ http://www.electricaleasy.com/p/electri

cal-machines.html

http://yourelectrichome.blogspot.in/

http://www.electricaleasy.com/p/electrical-machines.html


NPTEL Reference

• Electrical Machines II by Dr. Krishna Vasudevan & Prof. G. Sridhara RaoDepartment of Electrical Engineering , IIT Madras.

• Basic Electrical Technology by Prof. L. Umanand - IISc Bangalore video

BASICS OF ELECTRICAL MACHINES

Electrical Machine?Electrical machine is a device whichcan convert Mechanical energy into electrical

energy (Generators/alternators) Electrical energy into mechanical

energy (Motors) AC current from one voltage level to

other voltage level without changing its frequency (Transformers)

Presented by C.GOKUL,AP/EEE Velalar College of Engg & Tech , Erode

http://www.electricaleasy.com/2012/12/what-is-electrical-machine.html

Fundamental Principle..

Electrical Machines (irrespective of AC or DC) work on the fundamental principle of Faraday’s law of Electromagnetic Induction.

Faraday’s Law

Faraday’s Law of Electromagnetic Induction states that an EMF is induced in a coil when the magnetic flux linking this coil changes with time

or The EMF generated is proportional to

the rate at which flux is changed.d de Ndt dtψ ϕ

= − = −

Faraday’s Law – Illustration

Two forms of Induced EMF !

The effect is same if the magnet is moved and the coil is made stationery

We call it as statically induced EMF The previous case is referred to as

Dynamically induced EMF

Governing Rules

It becomes evident that there exists a relationship between mechanical energy, electrical energy and magnetic field.

These three can be combined and precisely put as governing rules each for generator and for motor

Fleming’s Right hand rule

For Generator

Fleming's Right hand rule(for Generator)

Fleming’s Left hand rule

For Motor

Fleming's left hand rule (for motors)

First finger - direction of magnetic field (N-S) Second finger - direction of current

(positive to negative) Thumb - movements of the wire

Maxwell’s Corkscrew rule

If the electric current is moving away from the observer, the direction of lines of force of the magnetic field surrounding the conductor is clockwise and that if the electric current is moving towards an observer, the direction of lines of force is anti-clockwise

Corkscrew (Screw driver) rule -Illustration

Coiling of Conductor

To augment the effect of flux, we coil the conductor as the flux lines aid each other when they are in the same direction and cancel each other when they are in the opposite direction

Many a times, conductor is coiled around a magnetic material as surrounding air weakens the flux

We refer the magnetic materialas armature core

Electromagnet

The magnetic property of current carrying conductor can be exploited to make the conductor act as a magnet – Electromagnet

This is useful because it is very difficult to find permanent magnets with such high field

Also permanent magnets are prone to ageing problems

AC Fundamentals

AC Fundamentals - continued

Whenever current passes through a conductor…

Opposition to flow of current Opposition to sudden change in current Opposition to sudden change in voltage Flux lines around the conductor

Inductive Effect

Reactance EMF Lenz Law

An induced current is always in such a direction as to oppose the motion or change causing it

Capacitive effect

( ) 1( ) ( )q tV t i t dtC C

= = ∫( ) ( )( ) dq t dv ti t C

dt dt⇒ = =

QCV

=

Resistive Network – Vector diagram

Inductive Network – Vector Diagram

Capacitive Network – Vector Diagram

Inductive & Capacitive effects -combined

Pure L & C networks – not at all possible!

R-L network

Pure L & C networks – not at all possible! – contd.

R-C network

Current & Flux

As already mentioned,As the current, so the flux

3 phase AC

Star and Delta

Star connection

3L ph

L ph

V V

I I

=

=

Delta Connection

3L ph

L ph

V V

I I

=

=

Maxwell's Right Hand Grip Rule

Right Handed Cork Screw Rule

Generators The Generator converts mechanical power into

electrical power.

Synchronous generators (Alternator) are constant speed generators.

The conversion of mechanical power into electrical power is done through a coupling field (magnetic field).

MagneticMechanical ElectricalInput Output

Electric Generator

GMechanicalEnergy

ElectricalEnergy

Stationary magnets - rotating magnets - electromagnets

Motor The Motor converts electrical power into

mechanical power.

Magnetic MechanicalElectrical

Input Output

MElectricalEnergy

MechanicalEnergy

Basic Construction Parts

Stator

Mechanical

Rotor

Armature

Electrical

Field

Rotating Part

Stationary Part

AC MACHINES Two categories:1.Synchronous Machines: Synchronous Generators(Alternator) Primary Source of Electrical Energy Synchronous Motor

2.Asynchronous Machines(Induction Machines)

UNIT-1Synchronous

Generator(Alternator)

UNIT-1 Syllabus

Synchronous Generators

Generator

Exciter

View of a two-pole round rotor generator and exciter. (Westinghouse)

Synchronous Machines

• Synchronous generators or alternators are used to convert mechanical power derived from steam, gas, or hydraulic-turbine to ac electric power

• Synchronous generators are the primary source of electrical energy we consume today

• Large ac power networks rely almost exclusively on synchronous generators

• Synchronous motors are built in large units compare to induction motors (Induction motors are cheaper for smaller ratings) and used for constant speed industrial drives

Construction

Basic parts of a synchronous generator:

• Rotor - dc excited winding • Stator - 3-phase winding in which the ac emf is generated

The manner in which the active parts of a synchronous machine are cooled determines its overall physical size and structure

Armature Windings (On Stator)• Armature windings connected are 3-phase and are

either star or delta connected• It is the stationary part of the machine and is built up of

sheet-steel laminations having slots on its inner periphery.

• The windings are 120 degrees apart and normally use distributed windings

Field Windings (on Rotor)• The field winding of a synchronous machine is always

energized with direct current

• Under steady state condition, the field or exciting current is given

Ir = Vf/Rf

Vf = Direct voltage applied to the field windingRf= Field winding Resistance

Rotor• Rotor is the rotating part of the machine

• Can be classified as: (a) Cylindrical Rotor and (b) Salient Pole rotor

• Large salient-pole rotors are made of laminated poles retaining the winding under the pole head.

Various Types of ROTOR

Salient-pole Rotor

Cylindrical or round rotor

1. Most hydraulic turbines have to turn at low speeds (between 50 and 300 r/min)

2. A large number of poles are required on the rotor

Hydrogenerator

TurbineHydro (water)

D ≈ 10 m

Non-uniform air-gap

N

S S

N

d-axis

q-axis

a. Salient-Pole Rotor

• Salient pole type rotor is used in low and medium speed alternators

• This type of rotor consists of large number of projected poles (called salient poles)

• Poles are also laminated to minimize the eddy current losses.

• This type of rotor are large in diameters and short in axial length.

Salient-Pole Synchronous Generator

Stator

L ≈ 10 m

D ≈ 1 mTurbine

Steam

Stator

Uniform air-gap

Stator winding

Rotor

Rotor winding

N

S

High speed 3600 r/min ⇒ 2-pole

1800 r/min ⇒ 4-pole

Direct-conductor cooling (using hydrogen or water as coolant)

Rating up to 2000 MVA

Turbogenerator

d-axis

q-axis

b. Cylindrical-Rotor(Non-Salient Pole)

• Cylindrical type rotors are used in high speed alternators (turbo alternators)

• This type of rotor consists of a smooth and solid steel cylinder having slots along its outer periphery.

• Field windings are placed in these slots.

Cylindrical-Rotor Synchronous Generator

Stator

Cylindrical rotor

Working of Alternator & frequency of Induced EMF

Working Principle• It works on the principle of Electromagnetic induction• In the synchronous generator field system is rotating and armature

winding is steady.• Its works on principle opposite to the DC generator• High voltage AC output coming from the armature terminal

Working Principle

• Armature Stator• Field Rotor• No commutator is

required No need for commutator because we need AC only

Every time a complete pair of poles crosses the conductor, the induced voltage goes through one complete cycle. Therefore, the generator frequency is given by

12060.

2pnnpf ==

Frequency of Induced EMF

N=Rotor speed in r.p.mP=number of rotor poles

f=frequency of induced EMF in Hz

No of cycles/revolution = No of pairs of poles = P/2No of revolutions/second = N/60No of cycles/second Frequency= (P/2)*(N/60)=PN/120

Advantages of stationary armature

• At high voltages, it easier to insulate stationary armature winding(30 kV or more)

• The high voltage output can be directly taken out from the stationary armature.

• Rotor is Field winding. So low dc voltage can be transferred safely

• Due to simple construction High speed of Rotating DC field is possible.


Winding Factors( Kp,Kd)

cos2

sin2

sin2

p

d

K

m

Km

α

β

β

=

=

Pitch factor (Kp) Consider 4 pole, 3 phase machine having 24

conductors Pole pitch = 24 / 4 = 6 slots If Coil Pitch or Coil Span = pole pitch, then it

is referred to as full-pitched winding If Coil Pitch < pole pitch, it is referred to as

short-pitched winding

Coil Span = 5 / 6 of pole pitch If falls short by 1 / 6 of pole pitch

or 180 / 6 = 30 degrees

This is done primarily to Save copper of end connections Improve the wave-form of the generated emf

(sine wave) Eliminate the high frequency harmonics

There is a disadvantage attached to it Total voltage around the coil gets reduced

because, the emf induced in the two sides of the coil is slightly out of phase

Due to that, their resultant vectorial sum is less than the arithmetic sum

This is denoted by a factor Pitch factor, Kp or Kc

Pitch factor – Kp

pVectorsumK

Arithmaticsum=

Pitch factor – contd.

Arithmatic sum


Vector sum



__

2 cos2

2

cos2

p

s

s

Vector sumKArithmatic sum

E

E

α

α

=

=

=

Pitch factor - Problem

Distribution factor (Kd) As we know, each phase consists of

conductors distributed in number of slots to form polar groups under each pole

The result is that the emf induced in the conductors constituting the polar group are not in phase rather differ by an angle equal to angular displacement of the slots

For a 3 phase machine with 36 conductors, 4 pole, no. of slots (conductors) / pole / phase is equal to 3

Each phase consists of 3 slots Angular displacement between any two adjacent

slots = 180 / 9 = 20 degrees If the 3 coils are bunched in 1 slot, emf induced is

equal to the arithmetic sum (3Es) Practically, in distributed winding, vector sum has to

be calculated Kd = Vector sum / Arithmetic sum

_ _ __ _ _d

emf with distributed windingKemf with concentrated winding

=

0 0180 180. _ _ _no of slots per pole n

β = =

For calculating Vector sum

2 sin2

2 sin2

sin2

sin2

d

d

mrK

m r

m

Km

β

β

β

β

=

=

Problem:Distribution factor /Breadth factor

EMF Equation of Alternator

Equation of Induced EMF

Average emf induced per conductor = dφ / dtHere, dφ = φP

If P is number of poles and flux / pole is φ Weber

dt = time for N revolution = 60 / N secondTherefore, Average emf = dφ / dt = φP / (60 / N)

60NPϕ

=

Equation of Induced EMF – contd.

We know, N = 120 f / PSubstituting, N we get Avg. emf per conductor = 2 f φ Volt If there are Z conductors / ph, then

Avg. emf induced / ph = 2 f φ Z Volt Ave emf induced (in turns) / ph = 4 f φ T Volt

Equation of Induced EMF – contd.

We know, RMS value / Avg. Value = 1.11 Therefore, RMS value of emf induced / ph = 1.11 (4 f φ T) V

= 4.44 f φ T Volt This is the actual value, but we have two other

factors coming in the picture, Kc and Kd These two reduces the emf induced

RMS value of emf induced = (Kd) (Kc) 4.44 f φ T Volt

Armature Reaction of Alternator

Armature Reaction

Main Flux Field Winding Secondary Flux Armature Winding Effect of Armature Flux on the Main Flux is

called Armature Reaction

Armature Reaction in alternator

I.) When load p.f. is unityII.) When load p.f. is zero laggingIII.) When load p.f. is zero leading

Armature Reaction in alternator

I.) When load p.f. is unity distorted but not weakened.- the average flux in the

air-gap practically remains unaltered.II.) When load p.f. is zero lagging the flux in the air-gap is weakened- the field

excitation will have to be increased to compensateIII.) When load p.f. is zero leading

the effect of armature reaction is wholly magnetizing- the field excitation will have to be reduced

1. Unity Power Factor Load

Consider a purely resistive load connected to the alternator, having unity power factor. As induced e.m.f. Eph drives a current of Iaph and load power factor is unity, Eph and Iph are in phase with each other.

If Φf is the main flux produced by the field winding responsible for producing Eph then Eph lags Φf by 90o .

Now current through armature Ia, produces the armature flux say Φa. So flux Φa and Ia are always in the same direction.

• Phase difference of 90o between the armature flux and the main flux• the two fluxes oppose each other on the left half of each pole while assist

each other on the right half of each pole.• Average flux in the air gap remains constant but its distribution gets distorted.

• Due to such distortion of the flux, there is small drop in the terminal voltage

2. Zero Lagging Power Factor Load

Consider a purely inductive load connected to the alternator, having zero lagging power factor.

Iaph driven by Eph lags Eph by 90o which is the power factor angle Φ.

Induced e.m.f. Eph lags main flux Φf by 90o while Φa is in the same direction as that of Ia.

the armature flux and the main flux are exactly in opposite direction to each other.

• As this effect causes reduction in the main flux, the terminal voltage drops. This drop in the terminal voltage is more than the drop corresponding to the unity p.f. load.

3. Zero Leading Power Factor Load

Consider a purely capacitive load connected to the alternator having zero leading power factor.

This means that armature current Iaph driven by Eph, leads Eph by 90o, which is the power factor angle Φ.

Induced e.m.f. Eph lags Φf by 90o while Iaph and Φa are always in the same direction.

the armature flux and the main field flux are in the same direction

• As this effect adds the flux to the main flux, greater e.m.f. gets induced in the armature. Hence there is increase in the terminal voltage for leading power factor loads.

Phasor Diagram for Synchronous

Generator/Alternator

Phasor Diagram of loaded Alternator

Ef which denotes excitation voltage Vt which denotes terminal voltage Ia which denotes the armature current θ which denotes the phase angle between Vt and Iaᴪ which denotes the angle between the Ef and Iaδ which denotes the angle between the Ef and Vt

ra which denotes the armature per phase resistanceTwo important points:

(1) If a machine is working as a synchronous generator then direction of Ia will be in phase to that of the Ef.

(2) Phasor Ef is always ahead of Vt.

Lagging PF Unity PF Leading PF

a. Alternator at Lagging PF Ef by first taking the component of the Vt in the

direction of Ia Component of Vt in the direction of Ia is Vtcosθ , Total voltage drop is (Vtcosθ+Iara) along the Ia. we can calculate the voltage drop along the direction

perpendicular to Ia. The total voltage drop perpendicular to Ia is

(Vtsinθ+IaXs). With the help of triangle BOD in the first phasor

diagram we can write the expression for Ef as

b. Alternator at Unity PF Ef by first taking the component of the Vt in

the direction of Ia. θ = 0 hence we have ᴪ=δ. With the help of triangle BOD in the second

phasor diagram we can directly write the expression for Ef as

c. Alternator at Leading PF Component in the direction of Ia is Vtcosθ. As the direction of Ia is same to that of the Vt thus

the total voltage drop is (Vtcosθ+Iara). Similarly we can write expression for the voltage

drop along the direction perpendicular to Ia. The total voltage drop comes out to be (Vtsinθ-IaXs). With the help of triangle BOD in the first phasor

diagram we can write the expression for Ef as

Determination of the parameters of the equivalent circuit from test data

The equivalent circuit of a synchronous generator that has been derived contains three quantities that must be determined in order to completely describe the behaviour of a real synchronous generator:

The saturation characteristic: relationship between If and φ (and therefore between If and Ef)The synchronous reactance, XsThe armature resistance, Ra

VOLTAGE REGULATION

Voltage regulation of an alternator is defined as the rise in terminal voltage of the machine expressed as a fraction of percentage of the initial voltage when specified load at a particular power factor is reduced to zero, the speed and excitation remaining unchanged.

Voltage Regulation

A convenient way to compare the voltagebehaviour of two generators is by theirvoltage regulation (VR). The VR of asynchronous generator at a given load,power factor, and at rated speed is definedas

%V

VEVR

fl

flnl 100×−

=

Voltage Regulation

Case 1: Lagging power factor:A generator operating at a lagging power factor has a positive voltage regulation.

Case 2: Unity power factor:A generator operating at a unity power factor has a small positive voltage regulation.

Case 3: Leading power factor:A generator operating at a leading power factor has a negative voltage regulation.

Voltage Regulation

This value may be readily determined from the phasor diagram for full load operation. If the regulation is excessive, automatic control of field current may be employed to maintain a nearly constant terminal voltage as load varies

Methods of Determination of voltage regulation

Methods of Determination of voltage regulation

Synchronous Impedance Method / E.M.F.MethodAmpere-turns method / M.M.F. methodZPF(Zero Power Factor) Method / PotierASA Method

1. Synchronous Impedance Method / E.M.F. Method

The method is also called E.M.F. method of determining the regulation. The method requires following data to calculate the regulation.

1. The armature resistance per phase (Ra).2. Open circuit characteristics which is the graph of open

circuit voltage against the field current. This is possible by conducting open circuit test on the alternator.

3. Short circuit characteristics which is the graph of short circuit current against field current. This is possible by conducting short circuit test on the alternator.

The alternator is coupled to a prime mover capableof driving the alternator at its synchronous speed.The armature is connected to the terminals of aswitch. The other terminals of the switch are shortcircuited through an ammeter. The voltmeter isconnected across the lines to measure the opencircuit voltage of the alternator.

The field winding is connected to a suitable d.c.supply with rheostat connected in series. The fieldexcitation i.e. field current can be varied with thehelp of this rheostat. The circuit diagram is shownin the Fig.

Circuit Diagram for OC & SC test

a. Open Circuit Test

Procedure to conduct this test is as follows :i) Start the prime mover and adjust the speed to the synchronous

speed of the alternator.ii) Keeping rheostat in the field circuit maximum, switch on the d.c.

supply.iii) The T.P.S.T switch in the armature circuit is kept open.iv) With the help of rheostat, field current is varied from its

minimum value to the rated value. Due to this, flux increasing the induced e.m.f. Hence voltmeter reading, which is measuring line value of open circuit voltage increases. For various values of field current, voltmeter readings are observed.

Open-circuit test Characteristics

The generator is turned at the rated speedThe terminals are disconnected from all loads, and

the field current is set to zero.Then the field current is gradually increased in

steps, and the terminal voltage is measured at each step along the way.

It is thus possible to obtain an open-circuit characteristic of a generator (Ef or Vt versus If) from this information

Connection for Open Circuit Test

Open-Circuit Characteristic

Short-circuit test

Adjust the field current to zero and short-circuit the terminals of the generator through a set of ammeters. Record the armature current Isc as the field

current is increased. Such a plot is called short-circuit

characteristic.

Short-circuit test

After completing the open circuit test observation, the fieldrheostat is brought to maximum position, reducing fieldcurrent to a minimum value.

The T.P.S.T switch is closed. As ammeter has negligibleresistance, the armature gets short circuited. Then the fieldexcitation is gradually increased till full load current isobtained through armature winding.

This can be observed on the ammeter connected in thearmature circuit. The graph of short circuit armaturecurrent against field current is plotted from the observationtable of short circuit test. This graph is called short circuitcharacteristics, S.C.C.

Short-circuit test

Adjust the field current to zero and short-circuit the terminals of the generator through a set of ammeters.

Record the armature current Isc as the field current is increased.

Such a plot is called short-circuit characteristic.

Connection for Short Circuit Test

Open and short circuit characteristic

Curve feature

The OCC will be nonlinear due to the saturation of the magnetic core at higher levels of field current. The SCC will be linear since the magnetic core does not saturate under short-circuit conditions.

Determination of Xs

For a particular field current IfA, the internal voltage Ef (=VA) could be found from the occ and the short-circuit current flow Isc,A could be found from the scc.

Then the synchronous reactance Xs could be obtained using

IfA

Ef or Vt (V) Air-gap line

OCC Isc (A)

SCC

If (A)

Vrated

VA Isc,B

Isc, A

IfB

( )scA

fAunsat,saunsat,s I

EVXRZ

==+= 22

22aunsat,sunsat,s RZX −=

scA

oc,t

scA

funsat,s I

VIE

X =≈

: Ra is known from the DC test.

Since Xs,unsat>>Ra,

Xs under saturated condition

( )scB

fratedsat,sasat,s I

EVXRZ

==+= 22

At V = Vrated,

22asat,ssat,s RZX −= : Ra is known from the DC test.

IfA


OCC Isc (A)

SCC

If (A)

Vrated

VAIsc,B

Isc, A

IfB

Advantages and Limitations of Synchronous Impedance Method

The value of synchronous impedance Zs for any load condition can be calculated. Hence regulation of the alternator at any load condition and load power factor can be determined. Actual load need not be connected to the alternator and hence method can be used for very high capacity alternators.

The main limitation of this method is that the method gives large values of synchronous reactance. This leads to high values of percentage regulation than the actual results. Hence this method is called pessimistic method

Equivalent circuit & phasor diagram under condition

Ia

Ef Vt=0

jXs Ra

+

+

EfVt=0

jIaXs

IaRa

Ia

Short-circuit Ratio Another parameter used to describe synchronous generators is the

short-circuit ratio (SCR). The SCR of a generator defined as the ratio of the field current required for the rated voltage at open circuit to the field current required for the rated armature current at short circuit. SCR is just the reciprocal of the per unit value of the saturated synchronous reactance calculated by

[ ].u.pinX

II

SCR

sat_s

Iscrated_f

Vrated_f

1=

=


OCCIsc (A)

SCC

If (A)

Vrated

Isc,rated

If_V rated If_Isc rated

Synchronous Generator Capability Curves

Synchronous generator capability curves are used to determine the stability of the generator at various points of operation. A particular capability curve generated in Lab VIEW for an apparent power of 50,000W is shown in Fig. The maximum prime-mover power is also reflected in it.

Capability Curve

2. MMF method (Ampere turns method)Tests: Conduct tests to find OCC (up to 125% of rated voltage) refer diagram EMF SCC (for rated current) refer diagram EMF

3. ZPF method (Potier method) Tests: Conduct tests to find OCC (up to 125% of rated voltage) refer diagram EMF SCC (for rated current) refer diagram EMF ZPF (for rated current and rated voltage) Armature Resistance (if required)


4. ASA method Tests: Conduct tests to find OCC (up to 125% of rated voltage) refer diagram EMF SCC (for rated current) refer diagram EMF ZPF (for rated current and rated voltage) Armature Resistance (if required)

Losses and Efficiency

The losses in synchronous generator include:1. Copper losses in

a) Armature b) Field windingc) The contacts between brushes

2. Core losses, Eddy current losses and Hysteresis losses

Losses

3. Friction and windage losses,the brush friction at the slip rings.

4. Stray load losses caused by eddy currents in the armature conductors and by additional core loss due to the distribution of magnetic field under load conditions.

synchronous generator power flow diagram

The three-phase synchronous generator power flow diagram

Synchronization & Parallel

operation of Alternator

Parallel operation of synchronous generators

There are several major advantages to operate generators in parallel:

• Several generators can supply a bigger load than one machine by itself.

• Having many generators increases the reliability of the power system.

• It allows one or more generators to be removed for shutdown or preventive maintenance.

Before connecting a generator in parallel with anothergenerator, it must be synchronized. A generator is said to besynchronized when it meets all the following conditions:

• The rms line voltages of the two generators must be equal.

• The two generators must have the same phase sequence.• The phase angles of the two a phases must be equal.• The oncoming generator frequency is equal to the

running system frequency.

Synchronization

Load

Generator 2

Generator 1

Switch

a

b

c

a/

b/

c/

Parallel operation of synchronous generators

Most of synchronous generators are operating in parallel with other synchronous generators to supply power to the same power system. Obvious advantages of this arrangement are:

1. Several generators can supply a bigger load;2. A failure of a single generator does not result in a total power loss to the load

increasing reliability of the power system;3. Individual generators may be removed from the power system for maintenance

without shutting down the load;4. A single generator not operating at near full load might be quite inefficient.

While having several generators in parallel, it is possible to turn off some of them when operating the rest at near full-load condition.

Conditions required for paralleling

A diagram shows that Generator 2 (oncoming generator) will be connected in parallel when the switch S1 is closed.However, closing the switch at an arbitrary moment can severely damage both generators!

If voltages are not exactly the same in both lines (i.e. in a and a’, b and b’ etc.), a very large current will flow when the switch is closed. Therefore, to avoid this, voltages coming from both generators must be exactly the same. Therefore, the following conditions must be met:

1. The rms line voltages of the two generators must be equal.2. The two generators must have the same phase sequence.3. The phase angles of two a phases must be equal.4. The frequency of the oncoming generator must be slightly higher than the

frequency of the running system.

Conditions required for paralleling

If the phase sequences are different, then even if one pair of voltages (phases a) are in phase, the other two pairs will be 1200 out of phase creating huge currents in these phases.

If the frequencies of the generators are different, a large power transient may occur until the generators stabilize at a common frequency. The frequencies of two machines must be very close to each other but not exactly equal. If frequencies differ by a small amount, the phase angles of the oncoming generator will change slowly with respect to the phase angles of the running system.If the angles between the voltages can be observed, it is possible to close the switch S1 when the machines are in phase.

General procedure for paralleling generators

When connecting the generator G2 to the running system, the following steps should be taken:

1. Adjust the field current of the oncoming generator to make its terminal voltage equal to the line voltage of the system (use a voltmeter).

2. Compare the phase sequences of the oncoming generator and the running system. This can be done by different ways:1) Connect a small induction motor to the terminals of the oncoming generator

and then to the terminals of the running system. If the motor rotates in the same direction, the phase sequence is the same;

2) Connect three light bulbs across the open terminals of the switch. As the phase changes between the two generators, light bulbs get brighter (large phase difference) or dimmer (small phase difference). If all three bulbs get bright and dark together, both generators have the same phase sequences.

General procedure for paralleling generators

If phase sequences are different, two of the conductors on the oncoming generator must be reversed.

3. The frequency of the oncoming generator is adjusted to be slightly higher than the system’s frequency.

4. Turn on the switch connecting G2 to the system when phase angles are equal.

The simplest way to determine the moment when two generators are in phase is by observing the same three light bulbs. When all three lights go out, the voltage across them is zero and, therefore, machines are in phase.

A more accurate way is to use a synchroscope – a meter measuring the difference in phase angles between two aphases. However, a synchroscope does not check the phase sequence since it only measures the phase difference in one phase.

The whole process is usually automated…

Synchronization

LoadGenerator

Rest of the power system

Generator

Xs1Ef1

Xs2Ef2

XsnEfn

Infinite busV, f are constantXs eq = 0

G

Concept of the infinite bus

When a synchronous generator is connected to a powersystem, the power system is often so large that nothing, theoperator of the generator does, will have much of an effecton the power system. An example of this situation is theconnection of a single generator to the power grid. Ourpower grid is so large that no reasonable action on the partof one generator can cause an observable change inoverall grid frequency. This idea is idealized in the conceptof an infinite bus. An infinite bus is a power system so largethat its voltage and frequency do not vary regardless ofhow much real or reactive power is drawn from or suppliedto it.

Steady-state power-angle characteristics

Active and reactive power-angle characteristics

• P>0: generator operation• P<0: motor operation• Positive Q: delivering inductive vars for a generator action or

receiving inductive vars for a motor action• Negaive Q: delivering capacitive vars for a generator action or

receiving capacitive vars for a motor action

PmPe, Qe

Vt

Fig. Synchronous generator connected to an infinite bus.


• The real and reactive power delivered by a synchronous generator or consumed by a synchronous motor can be expressed in terms of the terminal voltage Vt, generated voltage Ef, synchronous impedance Zs, and the power angle or torque angle δ.

• Referring to Fig. 8, it is convenient to adopt a convention that makes positive real power P and positive reactive power Q delivered by an overexcited generator.

• The generator action corresponds to positive value of δ, while the motor action corresponds to negative value of δ.

Pm Pe, Qe

Vt

The complex power output of the generator in volt-amperes per phase is given by

*at

_IVjQPS =+=

where:Vt = terminal voltage per phaseIa* = complex conjugate of the armature current per phase

Taking the terminal voltage as reference

0jVV tt

_

+=

the excitation( at stator in case of motor) or the generated voltage,( )δ+δ= sinjcosEE ff

_


Pm Pe, Qe

Vt


Pm Pe, Qe

Vt

and the armature current,

( )s

ftf

s

t

_

f

_

a

_

jXsinjEVcosE

jXVEI

δ+−δ=

−=

where Xs is the synchronous reactance per phase.

( )

s

tft

s

ft

s

tft

s

ft

s

ftft

*a

_t

_

XVcosEV

Q

&X

sinEVP

XVcosEV

jX

sinEV

jXsinjEVcosE

VIVjQPS

2

2

−δ=

δ=∴

−δ+

δ=

−

δ−−δ==+=


Pm Pe, Qe

Vt

s

tft

s

ft

XVcosEV

Q&X

sinEVP

2−δ=

δ=∴

• The above two equations for active and reactive powers hold good for cylindrical-rotor synchronous machines for negligible resistance

• To obtain the total power for a three-phase generator, the above equations should be multiplied by 3 when the voltages are line-to-neutral

• If the line-to-line magnitudes are used for the voltages, however, these equations give the total three-phase power

Steady-state power-angle or torque-angle characteristic of a cylindrical-rotor synchronous machine (with negligible

armature resistance).

+δ

Real power or torque

generator

motor

+π+π/2−π/2

0−π

Pull-out torque as a generator

Pull-out torque as a motor

−δ

Steady-state stability limit

Total three-phase power: δ= sinX

EVP

s

ft3

The above equation shows that the power produced by asynchronous generator depends on the angle δ between the Vt andEf. The maximum power that the generator can supply occurs whenδ=90o.

s

ft

XEV

P3

=

The maximum power indicated by this equation is called steady-statestability limit of the generator. If we try to exceed this limit (such as byadmitting more steam to the turbine), the rotor will accelerate and losesynchronism with the infinite bus. In practice, this condition is neverreached because the circuit breakers trip as soon as synchronism islost. We have to resynchronize the generator before it can again pickup the load. Normally, real generators never even come close to thelimit. Full-load torque angle of 15o to 20o are more typical of realmachines.

Pull-out torque

The maximum torque or pull-out torque per phase that a two-pole round-rotor synchronous motor can develop is

π

=ω

=

602 s

max

m

maxmax n

PPT

where ns is the synchronous speed of the motor in rpm

P

δ

P or Q

Q

Fig. Active and reactive power as a function of the internal angle

BLONDELS TWO REACTION THEORY


BLONDELS TWO REACTION THEORY

In case of cylindrical pole machines, the direct-axis and the quadrature axis mmfs act on the same magnetic circuits, hence they can be summed up as complexors. However, in a salient-pole machine, the two mmfs do not act on the same magnetic circuit.

The direct axis component Fad operates over a magnetic circuit identical with that of the field system, while the q-axis component Faq is applied across the interpole space, producing a flux distribution different from that of Fad or the Field mmf.

The Blondel's two reaction theory henceconsiders the results of the cross and direct-reaction components separately and if saturationis neglected, accounts for their different effectsby assigning to each an appropriate value forarmature-reaction "reactive" respectively Xaq andXad .

Considering the leakage reactance, the combined reactance values becomes

Xad = X + X ad and X sq = X aq

Xsq < Xsd as a given current component of the q-axis gives rise to a smaller flux due to the higher reluctance of the magnetic path.

• Let lq and Id be the q and d-axis components of the current I in the armature reference to the phasor diagram in Figure. We get the following relationships

• Iq= I cos (σ+θ) Ia = I cosφ

• Id = I sin (σ+ φ) Ir = I sinφ

I = √(Id2 + Iq2)= = √(Id2 + Ir2)

• where Ia and Ir are the active and reactive components of current I.

SLIP TEST

Slip Test (for salient pole machines only)

Short Circuit Transients for Synchronous

Generator

Short Circuit PhenomenonConsider a two pole elementary single phase alternator with concentrated stator winding as shown in Fig. 4. Consider a two pole elementary single phase alternator with concentrated stator winding as shown in Fig. 4.

The corresponding waveforms for stator and rotor currents are shown in the Fig

Let short circuit occurs at position of rotor shown in Fig. 4(a)when there are no stator linkages. After 1/4 Rev as shown Fig. 4(b), ittends to establish full normal linkage in stator winding. The statoropposes this by a current in the shown direction as to force the flux inthe leakage path. The rotor current must increase to maintain its fluxconstant. It reduces to normal at position (c) where stator current isagain reduces to zero. The waveform of stator current and field currentshown in the Fig. 5. changes totally if the position of rotor at the instantof short circuit is different. Thus the short circuit current is a function ofrelative position of stator and rotor.

Using the theorem of constant linkages a three phase shortcircuit can also be studied. After the instant of short circuit the fluxlinking with the stator will not change. A stationary image of main poleflux is produced in the stator. Thus a d.c. component of current iscarried by each phase.

The magnitude of d.c. component of current is different for eachphase as the instant on the voltage wave at which short circuit occurs isdifferent for each phase. The rotor tries to maintain its own poles

The rotor current is normal each time when rotor polesoccupy the position same as that during short circuit and thecurrent in the stator will be zero if the machine is previouslyunloaded. After one half cycle from this position the stator androtor poles are again coincident but the poles are opposite. Tomaintain the flux linkages constant, the current in rotor reaches toits peak value.

The stationary field produced by poles on the statorinduces a normal frequency emf in the rotor. Thus the rotorcurrent is fluctuating whose resultant a.c. component developsfundamental frequency flux which rotates and again produces inthe stator winding double frequency or second harmoniccurrents. Thus the waveform of transient current consists offundamental, a.c. and second harmonic components of currents.

Thus whenever short circuit occurs in three phase generatorthen the stator currents are distorted from pure sine wave andare similar to those obtained when an alternating voltage issuddenly applied to series R-L circuit.

Stator Currents during Short Circuit• If a generator having negligible resistance, excited and

running on no load is suddenly undergoing short circuit at itsterminals, then the emf induced in the stator winding is usedto circulate short circuit current through it. Initially thereactance to be taken into consideration is not thesynchronous reactance of the machine. The effect of armatureflux (reaction) is to reduce the main field flux.

• But the flux linking with stator and rotor can not changeinstantaneously because of the induction associated with thewindings. Thus at the short circuit instant, the armaturereaction is ineffective. It will not reduce the main flux. Thusthe synchronous reactance will not come into picture at themoment of short circuit. The only limiting factor for shortcircuit current at this instant is the leakage reactance.

After some time from the instant of short circuit, thearmature reaction slowly shows its effect and the alternator thenreaches to steady state. Thus the short circuit current reaches tohigh value for some time and then settles to steady value.

It can be seen that during the initial instant of short circuitis dependent on induced emf and leakage reactance which issimilar to the case which we have considered previously ofvoltage source suddenly applied to series R-L circuit. Theinstant in the cycle at which short occurs also affects the shortcircuit current. Near zero e.m.f. (or voltage) it has doublingeffect. The expressions that we have derived are applicable onlyduring initial conditions of short circuit as the induced emf alsoreduces after some tome because of increased armaturereaction.

The short circuit currents in the three phases during shortcircuit are as shown in the Fig(next slide)

Capability Curves of Synchronous Generators

• The rating of synchronous generators is specified in terms ofmaximum apparent power in KVA and MVA load at a specifiedpower factor (normally 80, 85 or 90 percent lagging) and voltage forwhich they are designed to operate under steady state conditions.This load is carried by the alternators continuously withoutoverheating. With the help of automatic voltage regulators theterminal voltage of the alternator is kept constant (normally within±5% of rated voltage).

• The power factor is also important factor that must be specified.This is because the alternator that is designed to operate at 0.95 p.f.lagging at rated load will require more field current when operate at0.85 p.f. lagging at rated load. More field current results inoverheating of the field system which is undesirable. For thiscompounding curves of the alternators can be drawn.

• If synchronous generator is supplying power at constantfrequency to a load whose power factor is constant then curveshowing variation of field current versus armature current whenconstant power factor load is varied is called compounding curve foralternator.

• To maintain the terminal voltage constant the lagging power factors require more field excitation that that required for leading power factors. Hence there is limitation on output given by exciter and current flowing in field coils because of lagging power factors.

• The ability of prime mover decides the active power output of the alternator which is limited to a value within the apparent power rating. The capability curve for synchronous generator specifies the bounds within which it can operate safely.

• The loading on generator should not exceed the generator rating as it may lead to heating of stator. The turbine rating is the limiting factor for MW loading. The operation of generator should be away from steady state stability limit (δ = 90o). The field current should not exceed its limiting value as it may cause rotor heating.

• All these considerations provides performance curves which are important in practical applications. A set of capability curves for an alternator is shown in Fig. 2. The effect of increased Hydrogen pressure is shown which increases the cooling.

• When the active power and voltage are fixed the allowable reactive power loading is limited by either armature or field winding heating. From the capability curve shown in Fig. 2, the maximum reactive power loadings can be obtained for different power loadings with the operation at rated voltage. From unity p.f. to rated p.f. (0.8 as shown in Fig. 2), the limiting factor is armature heating while for lower power factors field heating is limiting factor.

This fact can be derived as follows :• If the alternator is operating is constant terminal voltage and

armature current which the limiting value corresponding to heating then the operation of alternator is at constant value of apparent power as the apparent power is product of terminal voltage and current, both of which are constant.

• If P is per unit active power and Q is per unit reactive power then per unit apparent power is given by,

• Similarly, considering the alternator to be operating at constant terminal voltage and field current (hence E) is limited to a maximum value obtained by heating limits.

• Thus induced voltage E is given by,

If Ra is assumed to be zero then

The apparent power can be written as,

Substituting value of Īa obtained from (1) in equation (2),

Taking magnitudes,

• This equation also represents a circle with centre at (0, -Vt2/Xs). These two circles are

represents in the Fig. 3 (see next post as Fig. 1). The field heating and armature heating limitation on machine operation can be seen from this Fig. 3 (see next post as Fig.1).

• The rating of machine which consists of apparent power and power factor is specified as the point of intersection of these circles as shown in the Fig. 4. So that the machine operates safely.

UNIT-2SYNCHRONOUS MOTOR

Presented byC.GOKUL

AP/EEE

UNIT 2 Syllabus

Synchronous Motor

3 phase AC supply is given to the stator and mechanical energy is obtained from the rotor

Reverse of alternator operation However, field poles are given electrical

supply to excite the poles (electromagnets !) Rated between 150kW to 15MW with speeds

ranging from 150 to 1800 rpm. Constant speed motor

Rotating Magnetic Field (RMF)

Basics – Rotating Magnetic Field

When 3 phase supply is given to the stator winding, 3 phase current flows which produces 3 phase flux

The MMF wave of the stator will have rotating effect on the rotor

The effect of the field will be equal to that produced by a rotating pole

Rotating Magnetic Field (R.M.F) –contd.

RMF – contd.

RMF – contd.

( ) ( )( ) ( )

sin sin .......................( )sin 120 sin 120 ...................( )

sin 240 sin 240 ...................( )

R m m

Y m m

B m m

t at b

t c

φ φ ω φ θφ φ ω φ θ

φ φ ω φ θ

= =

= − = −

= − = −

RMF – contd.

Looking back at the waveform again, we see that at any instant, one waveform has zero magnitude and one has a positive value and the other, negative value

Let us consider at the following instances –0, 60, 120, 180 degrees

RMF – contd.

Case (i) φ = 0 (look at the waveform)

RMF – contd.

Simply substitute φ = 0 in equations a, b, c

( ) ( )

( ) ( )

sin sin 0 0

3sin 120 sin 0 1202

3sin 240 sin 0 2402

R m m

Y m m m

B m m m

φ φ θ φ

φ φ θ φ φ

φ φ θ φ φ

= = =

= − = − = −

= − = − = +

RMF – contd.

Case (i) - Phasor diagram

RMF – contd.

RMF – contd.

Case (ii) φ = 60 (look at the waveform)

RMF – contd.


( ) ( )

( ) ( )

3sin sin 602

3sin 120 sin 60 1202

sin 240 sin 60 240 0

R m m m

Y m m m

B m m

φ φ θ φ φ

φ φ θ φ φ

φ φ θ φ

= = =

= − = − = −

= − = − =

RMF – contd.

RMF – contd.

RMF – contd.

Case (iii) φ = 120 (look at the waveform)

RMF – contd.


( ) ( )

( ) ( )

3sin sin1202

sin 120 sin 120 120 0

3sin 240 sin 120 2402

R m m m

Y m m

R m m m

φ φ θ φ φ

φ φ θ φ

φ φ θ φ φ

= = =

= − = − =

= − = − = −

RMF – contd.

RMF – contd.

Case (iv) φ = 180 (look at the waveform)

RMF – contd.


( ) ( )

( ) ( )

sin sin180 0

3sin 120 sin 180 1202

3sin 240 sin 180 2402

R m m

Y m m m

B m m m

φ φ θ φ

φ φ θ φ φ

φ φ θ φ φ

= = =

= − = − =

= − = − = −

RMF – contd.

RMF – contd.

It is found that the resultant flux line is rotating at constant magnitude

This we refer as rotating field or revolving field

The speed at which it rotates will be at synchronous speed – Ns = (120 f / P )

Direction of rotation will be in the clockwise direction as shown in the previous slide

Principle of operation

Operation

We have a rotating field at the stator Rotor is another magnet If properly aligned (?!) these two magnets will

attract each other Since the stator field is rotating at

synchronous speed, it will carry the rotor magnet along with it due to attraction (magnetic locking)

Magnetic Locking - Illustration

Operation – contd.

Why - ?

It is true that magnetic locking will make the rotor run at synchronous speed

Locking cannot happen instantly in a machine (?)

This makes synchronous motors not self starting

Not self starting

Due to inertia

How to make Syn. Motor self starting

If the rotor is moved by external means (to overcome inertial force acting on it) then there is a chance for the motor to get started

Procedure to make SM self start

3 ph supply is given to the stator Motor is driven by external means Rotor is excited At an instant rotor poles will be locked with

the stator field and motor will run at syn. speed

Back EMF & V Curves ,

Inverted V Curves

EMF generation in a motor ? !

We call it as back emf Similar to generated emf in an alternator Rotor rotating at synchronous speed will

induce emf in the stationary armature conductors

The ac voltage applied has to overcome this back emf to circulate current through the armature winding

Back emf

As given, emf is proportional to flux

4.44b C dE K K fTφ=

Back emf

Slight deviation from the topic (?)

Coming back to Back emf

Increase in Load…

In a Synchronous motor with increase in load δ increases

Increase in Load, o.k – What about the speed ?

The speed of the Synchronous motor speed stays constant at synchronous speed even when the load is increased

Magnetic locking between the stator and rotor (stiffness of coupling) keeps the rotor run at synchronous speed

But when the angle of separation (δ) is 90, then stiffness (locking) is lost and the motor ceases to run

At constant load, varying the excitation…

Kindly see to it that

In all the cases discussed above, magnitude of current vector changes

Power factor changes But the product Icosφ would be constant so

that active power drawn by the machine remains constant

What actually happens ?

The resultant air gap flux is due to ac armature winding and dc field winding

If the field is sufficient enough to set up the constant air gap flux then the magnetizing armature current required from the ac source is zero – hence the machine operates at unity power factor – this field current is the normal field current or normal excitation


If the field current is less than the normal excitation – then the machine is under excited

This deficiency in flux must be made by the armature mmf – so the armature winding draws magnetizing current or lagging reactive MVA – leaving the machine to operate at lagging power factor


In case the field current is made more than its normal operation – then the machine is over excited

This excess flux must be neutralized by the armature mmf – so the machine draws demagnetizing current or leading reactive MVA – leaving the machine to operate at leading power factor

Better illustration

Better Illustration

Similarly,

Synchronous motor in pf improvement

This feature of synchronous motor makes it suitable for improving the power factor of the system

Motors are overexcited so that it draws leading current from the supply

The motor here is referred to as synchronous condenser

V - curves

Inverted V - curves

CIRCLE DIAGRAM

Circle Diagrams

This offers a quick graphical solution to many problems

Circle Diagrams – contd.

Excitation Circle diagram It gives the locus of armature current, as the

excitation voltage and load angle are varied

Excitation Circle Diagram

It is based on the voltage equation of a motor given by

It can be expressed as

t f a sV E I Z= +

fta

s s

EVIZ Z

= −

Excitation Circle Diagram – contd.

Each component in the above expression is a current component

It can be taken in such a way that they lag from their corresponding voltage component by power factor angle

fta

s s

EVIZ Z

= −



Same result can be obtained mathematically as follows

With Vt as referenceft

as s

EVIZ Z

= −

0 fta

s s

EVIZ Z

δφ φ

∠ −∠= −

∠ ∠


fta

s s

EVIZ Z

φ δ φ= ∠ − − ∠ − −

( ) ( ) ( )( )cos sin cos sinfta

s s

EVI j jZ Z

φ φ δ φ δ φ

= − − + − +

( ) ( )

Re

cos cos sin sinf ft ta

s s s s

arrangingE EV VI j

Z Z Z Zφ δ φ φ δ φ

= − + + − + +


( ) ( )2 2

2 cos cos sin sinf ft ta

s s s s

Magnitude

E EV VIZ Z Z Z

φ δ φ φ δ φ

= − + + − + +

( ) ( )2 2

2 2 cos cos sin sinf ft ta

s s s s

E EV VIZ Z Z Z

δ φ φ δ φ φ

= + − + + +

( ) ( )2 2

2 2 cos cos sin sin cos sin cos cos sin sinf ft ta

s s s s

E EV VIZ Z Z Z

δ φ δ φ φ δ φ δ φ φ

= + − − + +


( ) ( )2 2

2 2 cos cos sin sin cos sin cos cos sin sinf ft ta

s s s s

E EV VIZ Z Z Z

δ φ δ φ φ δ φ δ φ φ

= + − − + +

2 22 2 22 cos cos sin sin cos sin cos sin cos sinf ft ta

s s s s

E EV VIZ Z Z Z

δ φ δ φ φ δ φ φ δ φ

= + − − + +

2 22 2 22 cos cos cos sinf ft ta

s s s s

E EV VIZ Z Z Z

δ φ δ φ

= + − +

2 22 2 cosf ft ta

s s s s

E EV VIZ Z Z Z

δ

= + −


The above equation says that Vt / Zs is one side of a triangle, whose other side is given by Ef / Zs

The third side is given by Ia

2 22 2 cosf ft ta

s s s s

E EV VIZ Z Z Z

δ

= + −


Coming back to our diagram (kindly verify the sides)


In the diagram, if Vt is assumed constant, then Vt / Zs is a constant

Now, if Ef (the excitation) is fixed, Ef / Zs

vector and Ia vector follow the path of a circle as load is changed on the motor

This locus is referred to as Excitation circle Excitation circle defines the magnitude and

power factor of Ia and the load angle δ, for different shaft loads


Same old diagram

Power Circle Diagram

This again gives the locus of armature current, as the mechanical power developed and power factor is varied


Power output per phase is given as

P is the mechanical power developed including iron and mechanical losses

2cost a a aP V I I rφ= −


The equation can be written as, Dividing the whole equation by ra and

rearranging it, we get

2 cos 0ta a

a a

V PI Ir r

φ− + =

2 2 2 2cos sin cos 0ta a a

a a

V PI I Ir r

φ φ φ+ − + =


Subsitituting x = Ia sinφ and y = Ia cosφ, the equation becomes

This is equation of circle with

2 2 2 2cos sin cos 0ta a a

a a

V PI I Ir r

φ φ φ+ − + =

2 2 0t

a a

V Px y yr r

+ − + =

2

0, &2 2

t t

a a a

V V Pcentre radiusr r r

= = −



Alternatively, We know,

Adding Vt / 2 ra on either side we get,

2 cos 0ta a

a a

V PI Ir r

φ− + =

2 22 cos

2 2t t t

a aa a a a

V V VPI Ir r r r

φ

− + + =


Slight Modification, yields

2 22

Re ,

cos2 2

t t ta a

a a a a

arranging

V V V PI Ir r r r

φ

+ − = −

2 22 2 cos

2 2 2t t t

a aa a a a

V V V PI Ir r r r

φ

+ − = −


The above expression shows that

is one side of a triangle whose other two sides are Ia and Vt / 2ra seperated by φ

2 22 2 cos

2 2 2t t t

a aa a a a

V V V PI Ir r r r

φ

+ − = −

2

2t

a a

V Pr r

−


Going back to the power circle diagram

Power Circle Diagram - Inference

At Pmax, armature current is in phase with Vt/2ra, hence the power factor is unity

Magnitude of armature current is given by Vt/2ra

Power Circle Diagram - Inference

At Pmax, we know, radius of the power circle is zero

Substituting, radius = 0, we get2

max 02

t

a a

V Pr r

− =

2

max 4t

a

VPr

⇒ =

Power Circle Diagram- Inference

Maximum power input,

Efficiency is given by

2

,max cos .12 2

t tin t a t

a a

V VP V I Vr r

φ

⇒ = = =

( )( )

2max

2,max

/ 450%

/ 2t a

in t a

V rPP V r

η = = =


As we see, 50 % efficiency is too low a value for synchronous motor

At this efficiency, since the losses are about half of that of the input, temperature rise reaches the permissible limit

As such, maximum power output presented earlier cannot be met in practice


V – curves (again?!)

We know, excitation circle diagram shows locus of armature current as a function of excitation voltage

Power circle diagram shows locus of armature current as a function of power

When these two circles are super imposed…

V – curves – contd.

TORQUE EQUATION & POWER EQUATION

Power Developed by Synchronous Motor

Consider the phasor diagram


In a motor power developed can be given as

Looking at the phasor diagram again

cosm b aP E I ψ=


We need to manipulate the vector diagram to arrive at the expression


Torque Developed by Synchronous Motor

We know(e), T (2π Ns) = P if Ns is in rps So, T = P / (2π Ns) or T = P / (2π Ns) if Ns is in rpm

Maximum power developed

Condition for maximum power developed can be found by differentiating the power expression by δ and equating it to zero (as usual)

( )2

cos cosb bm

s s

E V EPZ Z

θ δ θ= − −

( )

,

sin 0m b

s

DifferentiatingdP E Vd Z

θ δδ

= − − =

Maximum power developed -condition

( )sin 0b

s

E VZ

θ δ− − =

( )sin 0θ δ− =

0θ δ⇒ − =

θ δ⇒ =


Substituting θ = δ, in the power expression, we get,

2

,max cosb bm

s s

E V EPZ Z

δ= −

2

,max cosb bm

s s

orE V EPZ Z

θ= −


If

Substituting, cos θ = Ra / Zs

,max

0a

bm

s

RE VPZ

≈

=

2

,maxb b a

ms s s

E V E RPZ Z Z

= −


2

,maxb b a

ms s s

E V E RPZ Z Z

= −

( )2,max

,

42

sb a m

a

SolvingZE V V R PR

= ± −

Maximum power developed –condition

As the equation says, Power developed depends on excitation

Differentiating with respect to Eb

( )2

cos cosb bm

s s

E V EPZ Z

θ δ θ= − −

( )2

cos cos 0m b b

b b s s

dP E V EddE dE Z Z

θ δ θ

= − − =


( )2

cos cos 0m b b

b b s s

dP E V EddE dE Z Z

θ δ θ

= − − =

2s

ba

VZER

=


This is the value of Eb which will make developed power to be maximum

The maximum power is given by substituting the condition (Eb) in Pm expression

2s

ba

VZER

=

2 2

,max 2 4ma a

V VPR R

= −

Operation of infinite bus

barsPresented by C.GOKUL,AP/EEE Velalar College of Engg & Tech , Erode

Operation of AC Generators in Parallelwith Large Power Systems

• Isolated synchronous generator supplying its own load is very rare (emergency generators)

• In general applications more than one generator operating in parallel to supply loads

• In Iran national grid hundreds of generators share the load on the system

• Advantages of generators operating in parallel:1- several generators can supply a larger load2- having many generators in parallel increase the

reliability of power system3- having many generators operating in parallel allows

one or more of them to be removed for shutdown & preventive maintenance

4- if only one generator employed & not operating near full load, it will be relatively inefficient


INFINITE BUS• When a Syn. Gen. connected to power system,

power sys. is so large that nothing operator of generator does, have much effect on pwr. sys.

• Example: connection of a single generator to a large power grid (i.e. Iran grid), no reasonable action on part of one generator can cause an observable change in overall grid frequency

• This idea belong to definition of “Infinite Bus”which is: a so large power system, that its voltage & frequency do not vary, (regardless of amount of real and reactive power load)


• When a syn. Gen. connected to a power system:

1-The real powerversus frequency characteristic of such a system

2-And the reactivepower-voltagecharacteristic


• Behavior of a generator connected to a large systemA generator connected in parallel with a large system as shown

• Frequency & voltage of all machines must be the same, their real power-frequency (& reactive power-voltage) characteristics plotted back to back


• Assume generator just been paralleled with infinite bus, generator will be “floating” on the line, supplying a small amount of real power and little or no reactive power

• Suppose generator paralleled, however its frequency being slightly lower than system’s operating frequency At this frequency power supplied by generator is less than system’s operating frequency, generator will consume energy and runs as motor


• In order that a generator comes on line andsupply power instead of consuming it, we should ensure that oncoming machine’s frequency is adjusted higher than running system’s frequency

• Many generators have “reverse-power trip” system

• And if such a generator ever starts to consume power it will be automatically disconnected from line

Starting Methods of Syn Motor

• As seen earlier, synchronous motor is not self starting. It is necessary to rotate the rotor at a speed very near to synchronous speed. This is possible by various method in practice. The various methods to start the synchronous motor are,

1. Using pony motors2. Using damper winding3. As a slip ring induction motor4. Using small d.c. machine coupled to it.

1. Using pony motors• In this method, the rotor is brought to the

synchronous speed with the help of some external device like small induction motor. Such an external device is called 'pony motor'.

• Once the rotor attains the synchronous speed, the d.c. excitation to the rotor is switched on. Once the synchronism is established pony motor is decoupled. The motor then continues to rotate as synchronous motor.

2. Using Damper Winding

3. As a Slip Ring Induction MotorRefer Unit 3 for detail understanding

4. Using Small D.C. Machine• Many a times, a large synchronous motor are provided

with a coupled d.c. machine. This machine is used as a d.c. motor to rotate the synchronous motor at a synchronous speed. Then the excitation to the rotor is provided. Once motor starts running as a synchronous motor, the same d.c. machine acts as a d.c. generator called exciter. The field of the synchronous motor is then excited by this exciter itself.

Current loci for constant power input, constant

excitation and constant power developed

Refer Book for detail study

Current loci for constant power input

Current loci for constant power developed(PM)

Current locus for constant Excitation

HUNTING

Natural frequency of oscillations

Refer Book

Damper windings

Refer Book for detail study

Synchronous motors are not self starting machines. Thesemachines are made self starting by providing a special winding in therotor poles, known as damper winding or squirrel cage windings. Thedamper winding consists of short circuited copper bars embedded in theface of the rotor poles

When an ac supply is provided to stator of a 3-phasesynchronous motor, stator winding produces rotating magnetic field.Due to the damper winding present in the rotor winding of thesynchronous motor, machine starts as induction motor (Inductionmachine works on the principle of induction. Damper windings insynchronous motor will carryout the same task of induction motor rotorwindings.

Therefore due to damper windings synchronous motor starts asinduction motor and continue to accelerate). The exciter for synchronousmotor moves along with rotor. When the motor attains about 95% of thesynchronous speed, the rotor windings is connected to exciter terminalsand the rotor is magnetically locked by the rotating magnetic field ofstator and it runs as a synchronous motor.

Functions of Damper Windings:• Damper windings helps the synchronous motor to start

on its own (self starting machine) by providing startingtorque

• By providing damper windings in the rotor ofsynchronous motor "Hunting of machine“ can besuppressed.

When there is change in load, excitation or change inother conditions of the systems rotor of the synchronousmotor will oscillate to and fro about an equilibriumposition. At times these oscillations becomes moreviolent and resulting in loss of synchronism of the motorand comes to halt.

Synchronous Condensers

• When synchronous motor is over excited it takes leading p.f. current. If synchronous motor is on no load, where load angle δ is very small and it is over excited (Eb > V) then power factor angle increases almost up to 90o. And motor runs with almost zero leading power factor condition.


• This characteristics is similar to a normal capacitor which takes leading power factor current. Hence over excited synchronous motor operating on no load condition is called as synchronous condenser or synchronous capacitor. This is the property due to which synchronous motor is used as a phase advancer or as power improvement device.

Disadvantage of Low Power Factor• In various industries, many machines are of induction motor

type. The lighting and heating loads are supplied through transformers. The induction motors and transformers draw lagging current from the supply. Hence the overall power factor is very low and lagging in nature.

• The power is given by,P = VI cosΦ .............. single phase

... I = P/(VcosΦ)

The high current due to low p.f. has following disadvantages :

1. For higher current, conductor size required is more which increases the cost.2. The p.f. is given by

cosΦ = Active power/ Apparent = (P in KW)/ (S in KVA)

Thus for fixed active power P, low p.f. demands large KVA rating

alternators and transformers. This increases the cost.3. Large current means more copper losses and poor efficiency.4. Large current causes large voltage drops in transmission lines, alternators and other equipments. This results into poor regulation.

Unit-3Three phase Induction

Motor

Presented ByC.GOKUL

AP/EEE

UNIT 3 Syllabus

Construction of Induction Motor

Types of Rotor


Principle of Operation


SLIP(s)

Compare Induction motor &

Transformer

Equivalent circuit

Losses & Efficiency

356

Losses - Summary

Efficiency (η) =Poutput

Pinput

357

Motor Torque

Tm =9.55 Pm

n

9.55 (1 – s) Pr

ns (1 – s)=

= 9.55 Pr / ns

Tm = 9.55 Pr / ns


358

I2R losses in the rotor

Pjr = s Pr

Pjr = rotor I2R losses [W]

s = slip

Pr = power transmitted to the rotor [W]

Mechanical PowerPm = Pr - Pjr

= Pr - s Pr

= (1 – s) Pr

Torque-Slip Characteristics

Condition for Maximum Torque

LOAD TEST

LOAD TEST ON THREE PHASE INDUCTION MOTOR

NO LOAD TEST

No Load Test or Running Light Test orOpen Circuit Test

This test gives1. Core loss2. F & W loss3. No load current I0

5. Ic, Rc, Iμ, Xm6. Mechanical faults, noise

Rated per voltage V0, with rated freq is given to stator.Motor is run at NO LOAD

STATOR

AI0

VV0

R

YB

ROTOR

N

W0

P0, I0 and V0 are recordedP0 = I0

2r1+Pc+Pfw

4. No load power factor

No load power factor is small, 0.05 to 0.15

1. Ic=I0cosθ0 2. Iμ=I0sinθ0

3.

On No load, Motor runs near to syn speedSo, s ≈ zero 1/s=α or open circuit

4.

00

00 IV

PθC =os

)(, 11000c

0c jxrIVE

IER +−==

µIEX 0

m =

r1

r2/s

jx1 jx2I2

jXmRc

I0

I0

Ic IΦ

V0

open

cir

cuit

provided x1 is known

The F & W loss Pfw, can be obtained from thistest.

Vary input voltage and note input power

Input Power

Input Voltage

Pfw

Thus Pc=P0 - I02r1 - Pfw


BLOCKED ROTOR TEST

Rotor is blocked, Speed = 0, slip = 1Blocked Rotor test or Short Circuit Test

AIsc

V Vsc

R

YB

N

Wsc

3-ph Variac

I M

Rotor is blocked or held stationary bybelt pulley or by hand

Low voltage is applied upto rated stator current

Voltage Vsc, Current Isc and Power Psc are measured.

Mechanical loss =0

Rc and Xm >> r2+jx2

Therefore, Zsc = Vsc / Isc

=Rsc+jXsc

This test gives copper loss

Since slip is 1, secondary is short circuited

jx1 jx2

jXmRc

r1 Isc

I0

Ic IΦ

Vsc

r2

−

ss1r2

Core loss negligible

Hence omitted scscIVPcosθ sc

sc = =0.8 to 0.9

Class of motor x1 x2

= r1+r2

1. Class A (normal Tst and Ist) 0.5 0.5

For wound rotor motor, x1 = x2 = Xsc /2

Rsc= Psc/Isc2

22scscsc RZX −=

r2= Rsc – r1

21 xx +=

For squirrel cage motor,

2. Class B (normal Tst and low Ist) 0.4 0.63. Class C (high Tst and low Ist) 0.3 0.7

4. Class D (high Tst and high slip) 0.5 0.5

CIRCLE DIAGRAM

But the advantage of circle diagram is that

torque and slip can be known from circle diagram

The circle diagram is constructed with the help of

Graphical representationThe equivalent ckt., operating ch. can be obtained

by computer quickly and accurately

1. No load test (I0 & θ0)

2. Blocked rotor test (Isc & θsc)

Circle Diagram of Ind Motor

extremities or Limits of stator current, Power,

x

y

I0θ0

Isc

θsc

1. Draw x and y axes(V1 on y axis)2. Draw I0 and Isc(=V1/Zsc)3. Draw parallel line to x axis from I0.

This line indicates constant loss vertically

V1

Line I0Isc is output line

4. Join I0 and Isc

Output line

O

x

y

I0θ0

θsc

C

Output line

L1

T

V1

5. Draw perpendicular bisector to output line6. Draw circle with C as a centre7. Draw perpendicular from Isc on x axis..8. Divide IscL1 in such a way that. Loss Cu Stator

Loss Cu Rotorr'r

LTTI

1

2

1

sc ==

Isc

L2O

x

y

I0θ0

θsc

C

Output line

9. Join I0T. This is called as Torque Line. 10. Suppose 1cm=Xamp, so 1cm=V1.X= power scale

Rated output power/V1X = Total cm for rated o/p power

Torque line

V1

Total cm for rated output power=IscR

Isc

T

R

L1

L2O

rated output power

x

y

I0θ0

θsc

C

Output line

11. From R, draw line parallel to output line crossing at P & P’. P is operating point

Torque line

V1Isc

T

R

P

P’

L2’L1’T’O’

L1

L2

12. Join O and P. Cosθ1 is operating pf.

θ1

Lebel O’, T’ , L1’ and L2’13. From P draw perpendicular on x axis

O

x

y

I0θ0

θsc

C

Output line

14. Determine the following 1. Constant Losses and copper losses

=Core loss + F & W loss

Torque line

V1Isc

T

R

P

P’

L2’L1’T’O’

L1

L2

L1L2=L1’L2’=constant lossesα no load current I0

θ1

O

x

y

I0θ0

θsc

C

Output line

At standstill, input power = IscL2 L1L2=Constant Loss

Torque line

V1Isc

T

R

P

P’

L2’L1’T’O’

L1

L2

Constant loss= Stator core loss +rotor core loss (f)F & W loss=0

θ1

O

x

y

I0θ0

θsc

C

Output line

Torque line

V1Isc

T

R

P

P’

L2’L1’T’O’

L1

L2

At operating point P, input power = PL2’,

θ1

L1’L2’=Constant LossConstant loss = Stator core loss + F & W loss

Rotor core loss ≈ 0 (sf)Thus L1L2=L1’L2’= Constant loss

O

x

y

I0θ0

θsc

C

Output line

At standstill, Stator Cu loss=TL1

Torque line

V1Isc

T

R

P

P’

L2’L1’T’O’

L1

L2

rotor Cu loss = IscTAt P, stator Cu loss =T’L1’ and

θ1

rotor Cu loss = O’T’

O

x

y

I0θ0

θsc

C

Output line

Torque line

V1Isc

T

R

P

P’

L2’L1’T’O’

L1

L22. Output Power and Torque

θ1

Output Power = O’P The gap betn output line and circle is OUTPUT Power.At I0, o/p=0, at Isc, o/p=0

O”

Pmax

Max output power=PmaxO”Slip1 0

Ns0 Speed

Pmax

T”L1”L2”O

x

y

I0θ0

θsc

C

Output line

Torque line

V1Isc

T

R

P

P’

L2’L1’T’O’

L1

L2

θ1

Output Torque = T’P The gap betn torque line and circle is OUTPUT torque.At I0, torque=0, but at

Isc, torque=T Isc

Pmax

=Starting torque

Tmax

Max output torque=TmaxT”’

2. Output Power and Torque

Slip1 0Ns0 Speed

Tmax

O”T”L1”L2”

O”’

T”’L1”’L2”’O

x

y

I0θ0

θsc

C

Output line

Torque line

V1Isc

T

R

P

P’

L2’L1’T’O’

L1

L2

θ1

Max Power and Max Torque are not occurring at same timeContradiction to max power transfer theorem

Pmax

Tmax

2. Output Power and Torque

O”T”L1”L2”

O”’

T”’L1”’L2”’O

x

y

I0θ0

θsc

C

Output line

Torque line

V1Isc

T

R

P

P’

L2’L1’T’O’

L1

L2

θ1

Pmax

Tmax

O”T”L1”L2”

O”’

T”’L1”’L2”’

Air gap power Pg = Input power – Stator Cu loss- core loss=PL2’-T’L1’-L1’L2’

3. Slip, Power factor and Efficiency

= PT’

s = rotor Cu loss/Pg =O’T’/PT’ """

TPTOsmp

max

='"'"'"

TTTOsmt

max

=

O

x

y

I0θ0

θsc

C

Output line

Torque line

V1Isc

T

R

P

P’

L2’L1’T’O’

L1

L2

θ1

Pmax

Tmax

O”T”L1”L2”

O”’

T”’L1”’L2”’

3. Slip, Power factor and EfficiencyO

Power factor cosθ1 = PL2’/OP

PO’/PL2’Efficiency=

x

y

I0θ0

θsc

C

Output line

Torque line

V1Isc

T

R

P

P’

L2’L1’T’O’

L1

L2

θ1

s=0The gap betn circle and T & s=α is braking torque

O”

Pmax

Tmax

T”

4. Braking TorqueO

s=1

s=α

braking torque

Slip 0Ns0 Speed

Te

1α

x

y

I0θ0

θsc

C

Output line

Torque line

V1Isc

T

R

P

P’

L2’L1’T’O’

L1

L2

θ1

s=0

O”

Pmax

Tmax

T”

5. Induction GeneratorO

s=1

s=α

braking torque

x

y

I0θ0

θsc

C

Output line

Torque line

V1Isc

T

R

P

P’

L2’L1’T’O’

L1

L2

θ1

s=0

(Generator)

O”

Pmax

Tmax

T”


s=1

s=α

braking torque

s= -veθG

G

OG=Gen CurrentO’G=Mech I/p

L2’G=Active powerOL2’=reactive powerPGmax

x

y

I0θ0

θsc

C

Output line

Torque line

V1Isc

T

R

P

P’

L2’L1’T’O’

L1

L2

θ1

s=0

(Generator)

O”

Pmax

Tmax

T”


s=1

s=α

braking torque

s= -veθG

G

OG=Gen CurrentO’G=Mech I/p

L2’G=Active powerOL2’=reactive powerPGmax

Slip 0 -1

Speed2Ns

Ns0Slip

Speed

Te

1α

CIRCLE DIAGRAM OF AN INDUCTION MOTOR- Summary

H

T

Fig. 3.3

Separation of Losses


SEPARATION OF NO LOAD LOSSESThe separation of core loss and mechanical loss (windage and friction) can be obtained byno load test conducted from variable voltage, rated frequency supply. Step by step reducethe voltage till the machine slip suddenly start to increase and the motor tends to rest(stall). The core loss decrease almost square of the voltage and windage and friction lossremains almost constant. Plot the curve between applied voltage (V) and power (Po),extended to V=0 which gives mechanical loss.

Mechanical loss will be obtained from graphMagnetic loss + mechanical loss = output powerTherefore., magnetic loss = output power – mechanical loss

Formulae for calculating the equivalent circuit parameters:Z0 = Voc /(Ioc / √3) R0 = Woc / (Ioc) 2X0 = √[( Z0)2 - (R0)2ϕ0 = cos-1 [Woc / (√3 * Voc * Ioc )] RBR = Wsc / (Isc)2

ZBR = Vsc / (Isc/ √3)XBR = √[( ZBR)2 - (RBR)2]RiWF – Resistance accounting for rotational lossesR1 = 1.2 * stator winding resistance (dc)Pr = Woc – Ioc

2 * R1 (since Pr = P0 – 3 * (Ioc / √3)2 * R1)RiWF = Voc

2 / PrXm – Magnetizing reactance IiWF = Voc / RiwfIm = (Ioc

2 - IiWF2)1/2

Xm = Voc / Im

Equivalent Circuit:

Double cage Induction Motors

DOUBLE CAGE ROTOR

Double Cage Rotor has two independent cages on the same rotor slots, one inside the other for the production of high starting torque. The outer cage (alloy) in the rotor has high resistance and low reactance which is used for starting purpose. The inner cage (copper) has a low resistance and high reactance which is used for running purpose. The constructional arrangement and torque-speed characteristics as shown in fig. 3.5.

Advantages:

High starting torque.

Low I2R loss under running conditions and high efficiency.

Fig. 3.5

Double Cage construction

Torque-Slip CharacteristicsSlip

Equivalent Circuit:

If the magnetising current is neglected, then the equivalent circuit is reduced to

Rotor

‘ ‘

Induction Generators

Principle of operationInduction generators and motors produce electrical power when their rotor is rotated faster than the synchronous speed. For a four-pole motor operating on a 50 Hz will have synchronous speed equal to 1500 rpm.In normal motor operation, stator flux rotation is faster than the rotor rotation. This is causing stator flux to induce rotor currents, which create rotor flux with magnetic polarity opposite to stator. In this way, rotor is dragged along behind stator flux, by value equal to slip.In generator operation, a prime mover (turbine, engine) is driving the rotor above the synchronous speed. Stator flux still induces currents in the rotor, but since the opposing rotor flux is now cutting the stator coils, active current is produced in stator coils and motor is now operating as a generator and sending power back to the electrical grid.

INDUCTION GENERATOR

Fig. 3.4 current Locus for Induction Machine

a. Sub-synchronous (motor) b. Super-synchronous (generator)

Fig.3.5 Phasor DiagramFig. 3.6 Torque-Slip Characteristics

When the machine runs as induction generator, the vector diagram shown in fig.3.5. This is possible only if the machine is mechanically driven above the synchronous speed.

OA-no load currentAB-stator current to overcome rotor mmf

OB-total stator current

Fig.3.4b the point P in the lower half of the circle shows operating point as an induction generator.

PT-stator electrical outputST-Core, friction and windage lossesRS-Stator copper lossQR-Rotor copper lossPQ-Mechanical inputPR-Rotor input

Slip

Efficiency

PRQR

inputrotorlosscopperrotor

==

PQPT

inputoutput

==

The torque-slip curve is shown in fig.3.6.Torque will become zero at synchronous speed. If the speed increases above the synchronous speed, the slip will be negative.

Induction generator differs from the synchronous generator as Dc current excitation is not required. Synchronisation is not required.

Advantages: It does not hunt or drop out of synchronism Simple in construction Cheaper in cost Easy maintenance Induction regulators provide a constant voltage adjustment depending on the loading of the lines.Disadvantages: Cannot be operated independently. Deliver only leading current. Dangerously high voltages may occur over long transmission lines if the synchronous machines at the far end become disconnected and the line capacitance excites the induction machines. The induction generator is not helpful in system stability.Applications: For installation in small power stations where it can be operated in parallel and feeding into a common mains without attendant. For braking purpose in railway work.

Synchronous Induction Motor

SYNCHRONOUS INDUCTION MOTORIt is possible to make the slip ring induction motor to run at synchronous speed when its secondary winding is fed from a dc source. Such motors are then called as synchronous induction motor.

3Φ

Supply

Stator

Fig. 3.3

Rotor connections for dc excitation:

Heating will always occur with normal three phase rotor winding as in fig.3.4. The two phase windings (e and f) gives uniform heating but produce large harmonics and noise. In those machines primary chording is commonly employed to reduce the effect of harmonics.

The synchronous induction motor is generally built for outputs greater than 30HP because of its higher cost of the dc exciter. These motors are employed in applications where a constant speed is desirable such as compressors, fans, pumps, etc., If load torque is high and the machines goes out of synchronism, it continues to run as an induction motor. As soon as the load torque falls sufficiently low, the machines will automatically synchronize.

Fig 3.4

Advantages:

It will start and synchronise itself against heavy loads.

No separate damper winding is required.

The exciter may be small unit due to smaller air-gap.

Problems in Induction

Motors

Example 5.1 A 3-phase, 460 V, 100 hp, 60 Hz, four-pole induction machine delivers rated output power at a slip of 0.05. Determine the:

(a) Synchronous speed and motor speed.(b) Speed of the rotating air gap field.(c) Frequency of the rotor circuit.(d) Slip rpm.(e) Speed of the rotor field relative to the (i) rotor structure. (ii) Stator structure. (iii) Stator rotating field.(f) Rotor induced voltage at the operating speed, if the stator-to-rotor turns ratio is 1 :

0.5.

Solution:

rpmp

fns 1800

460*120120

===

( ) ( ) rpmnsn s 17101800*05.011 =−=−=(b) 1800 (same as synchronous speed)

Example 4.2 A no-load test conducted on a 30 hp, 835 r/min, 440 V, 3-phase, 60 Hz squirrel-cage induction motor yielded the following results:No-load voltage (line-to-line): 440 VNo-load current: 14 ANo-load power: 1470 WResistance measured between two terminals: 0.5 ΩThe locked-rotor test, conducted at reduced volt-age, gave the following results:Locked-rotor voltage (line-to-line): 163 VLocked-rotor power: 7200 WLocked-rotor current: 60 ADetermine the equivalent circuit of the motor.

Solution:Assuming the stator windings are connected in way, the resistance per phase is:

Ω== 25.02/5.01RFrom the no-load test:

PhaseVVV LL /2543

44031 ===

Ω=== 143.1814254

1

1

IVZNL

Ω=== 5.214*3

14703 22

1IP

R NLNL

97.175.2143.18 2222 =−=−= NLNLNL RZX

Ω==+ 97.171 NLm XXX

Ω=== 6667.060*3

7200

3 221 BL

BLBL

I

PR

From the blocked-rotor test

The blocked-rotor reactance is:

( ) Ω=−=−= 42.16667.05685.1 2222BLBLBL RZX

Ω=′+≅ 42.121 XXX BL

Ω=′=∴ 71.021 XX

Ω=−=−= 26.1771.097.171XXX NLmΩ=−=−= 4167.025.06667.01RRR BL

Ω=

+

=

+′=′∴ 4517.04167.0*

26.1726.1771.0 22

22 R

XXXR

m

m

Example 5.3 The following test results are obtained from a three-phase 60 hp, 2200 V, six-pole, 60 Hz squirrel-cage induction motor.

(1) No-load test:Supply frequency = 60 Hz, Line voltage = 2200 VLine current = 4.5 A, Input power = 1600 W(2) Blocked-rotor test:Frequency = 15 Hz, Line voltage = 270 VLine current = 25 A, Input power = 9000 W(3) Average DC resistance per stator phase: 2.8 Ω(a) Determine the no-load rotational loss.(b) Determine the parameters of the IEEE-recommended equivalent circuit(c) Determine the parameters (Vth, Rth, Xth) for the Thevenin equivalent circuit of

Fig.5.16.

PhaseVV /2.12703

22001 == Ω=== 27.282

5.42.1270

1

1

IVZNL

Ω=== 34.265.4*3

16003 22

1IPR NL

NL

(a) No-Load equivalent Circuit (b) Locked rotor equivalent circuit

Ω=−=−= 28134.2627.282 2222NLNLNL RZX

Ω==+ 2811 NLm XXX

=281.0 Ω.

Ω=== 8.425*3

90003 22

1IPR BL

BL

Ω=−=−=′ 28.28.412 RRR BL

impedance at 15 Hz is:Ω=== 24.6

25*3270

1

1

IVZBL

The blocked-rotor reactance at 15 Hz is ( ) Ω=−= 98.38.424.6 22BLX

Its value at 60 Hz is Ω== 92.151560*98.3BLX

21 XXX BL ′+≅

Ω==′=∴ 96.7292.15

21 XX at 60 Hz

Ω=−= 04.27396.7281mXΩ=−=−= 28.28.41RRR BL

Ω=

+

=′ 12.2204.273

04.27396.7 2

2R

)c (

11 97.004.27396.7

04.273 VVVth =+

≅

Ω==≅ 63.28.2*97.097.0 21

2 RRth

Ω=≅ 96.71XX th

Example 4.4 A three-phase, 460 V, 1740 rpm, 60 Hz, four-pole

wound-rotor induction motor has the following parameters per

phase:

1R = 0.25 Ω, 2.02 =′R Ω, 5.021 =′= XX Ω, 30=mX Ω

The rotational losses are 1700 watts. With the rotor terminals

short-circuited, find

(a) (i) Starting current when started direct on full voltage.

(ii) Starting torque.

(b) (i) Full-load slip.

(ii) Full-load current.

(iii) Ratio of starting current to full-load current.

(iv) Full-load power factor.

(v) Full-load torque.

(iv) Internal efficiency and motor efficiency at full load.

(c) (i) Slip at which maximum torque is developed.

(ii) Maximum torque developed.

(d) How much external resistance per phase should be

connected in the rotor circuit so that maximum torque occurs at

start?

=163.11 N.m

%5.87100*4.320223.28022

==motorη

( ) ( ) %7.96100*0333.01100*1int =−=−= sernalη

(c) (i)

(c) (ii)

Note that for parts (a) and (b) it is not necessary to use Thevenin equivalent circuit. Calculation can be based on the equivalent circuit of Fig.5.15 as follows:

A three-phase, 460 V, 60 Hz, six-pole wound-rotor induction motor

drives a constant load of 100 N - m at a speed of 1140 rpm when

the rotor terminals are short-circuited. It is required to reduce the

speed of the motor to 1000 rpm by inserting resistances in the

rotor circuit. Determine the value of the resistance if the rotor

winding resistance per phase is 0.2 ohms. Neglect rotational

losses. The stator-to-rotor turns ratio is unity.

Example The following test results are obtained from three

phase 100hp,460 V, eight pole star connected induction machine

No-load test : 460 V, 60 Hz, 40 A, 4.2 kW. Blocked rotor test is

100V, 60Hz, 140A 8kW. Average DC resistor between two stator

terminals is 0.152 Ω

(a) Determine the parameters of the equivalent circuit.

(b) The motor is connected to 3ϕ , 460 V, 60 Hz supply and runs

at 873 rpm. Determine the input current, input power, air

gap power, rotor cupper loss, mechanical power developed,

output power and efficiency of the motor.

(c) Determine the speed of the rotor field relative to stator

structure and stator rotating field

Solution: From no load test:

( ) Ω== 64.640

3/460NLZa

Ω=== 875.040*3

4200*3 22

1IPR NL

NL

Ω=−= 58.6875.064.6 22NLX

Ω=+ 58.61 mXXFrom blocked rotor test:

Ω== 136.0140*3

80002BLR Ω== 076.0

2152.0

1R

Ω== 412.0140

3/100BLZ

Ω=−= 389.0136.0412.0 22BLX

Ω==′= 1945.02389.0

21 XX3855.61945.058.6 =−=mX

Ω=−=−= 06.0076.0136.01RRR BL

0637.006.0*3855.6

3855.61945.0 2

2 =

+

=′R

0.076 j0.195

j6.386

j0.195

s0637.0

Ω Ω

Ω

Ω

389.021 =′+ XX

( ) rpmP

fnb s 9008

60*120120===

03.0900

873900=

−=

−=

s

s

nnns

123.203.0

0637.02 ==′

sR

Input impedance ( )( )( ) Ω∠=

+++

++= oj

jjjZ 16.27121.2195.0386.6123.2195.0123.2386.6195.0076.01

o

ZVI 16.2722.125

16.2712.23/460

1

11 −∠=

∠==

Input power:

( ) kWP oin 767.8816.27cos22.125*

3460*3 ==

Stator CU losses:

kWPst 575.3076.0*22.125*3 2 ==Air gap power kWPag 192.85575.3767.88 =−=

Rotor CU losses kWsPP ag 556.2192.85*03.02 ===

Mechanical power developed:

( ) ( ) kWPsP agmech 636.82192.85*03.011 =−=−=

rotmechout PPP −=

From no load test: WRIPP NLrot 2.3835076.0*40*34200*3 21

21 =−=−=

kWPout 8.782.383510*636.82 3 =−=

%77.88100*767.88

8.78100* ===in

out

PPη

Example A three phase, 460 V 1450 rpm, 50 Hz, four pole

wound rotor induction motor has the following parameters per

phase ( 1R =0.2Ω, 2R′ =0.18 Ω, 21 XX ′= =0.2Ω, mX =40Ω). The

rotational losses are 1500 W. Find,

(a) Starting current when started direct on full load voltage.

Also find starting torque.

(b) (b) Slip, current, power factor, load torque and efficiency

at full load conditions.

(c) Maximum torque and slip at which maximum torque will

be developed.

(d) How much external resistance per phase should be

connected in the rotor circuit so that maximum torque

occurs at start?

phaseVV /6.2653

4601 ==

( )Ω∠=

++

++= oj

jjZ 59.4655.02.4018.0

2.018.0*402.02.01

Ω−∠=∠

== oost I

VI 3.4691.48259.4655.0

6.2651

1

0333.01500

14501500=

−=s

4.50333.018.02 ==

′s

R

( )Ω∠=

++

++= oj

jjjZ 83.10959.44.454.5

2.04.5*402.02.01

AI ooFL 83.1056.53

83.10959.46.265

1 −∠=∠

=

Then the power factor is: 9822.083.10cos =o lag.

.sec/08.1572*60

1500 radsys == πω

( )( ) V

jjV o

th 285.0275.2642.402.0

40*6.265∠=

+=

Then,

( ) 2.0198.0285.45281432.02.402.0

2.02.0*40 jj

jjZ oth +=∠=

++

= Ω

( )( ) ( )

NmT 68.2282.02.04.5198.0*08.157

4.5*275.264*322

2=

+++=∴

Then, WTP sysag 1.3592108.157*68.228* === ω

Then, WsPP ag 11971.35921*0333.02 ===

And, ( ) WPsP agm 7.347231 =−=

Then, WPPP rotmout 7.3322315007.34723 =−=−=

WPin 419179822.0*56.53*6.265*3 ==

Then, %26.7941914

7.33223===

in

out

PPη

( )( )( )[ ] NmTm 56.862

2.02.0198.0198.05.188*2

275.264*32/122

2=

+++=∴

( )[ ] 4033.02.02.0198.0

18.02/122max

=++

=Ts

(d) ( )[ ] 2/1222

2.02.0198.01

max++

′+′== ext

TRRs

Then, 446323.02 =′+′ extRR

Then, 26632.018.0446323.0 =−=′extR Ω


Example 5.6 The rotor current at start of a three-phase, 460 volt,

1710 rpm, 60 Hz, four pole, squirrel-cage induction motor is six

times the rotor current at full load.

(a) Determine the starting torque as percent of full load torque.

(b) Determine the slip and speed at which the motor develops

maximum torque.

(c) Determine the maximum torque developed by the motor as

percent of full load torque.

Note that the equivalent circuit parameters are not given. Therefore equivalent circuit parameters cannot be used directly for computation.(a) The synchronous speed is

sRI

sRITsyn

2222

22 αω

=

Example 4.9 A 4 pole 50 Hz 20 hp motor has, at rated voltage

and frequency a starting torque of 150% and a maximum torque of

200 % of full load torque. Determine (i) full load speed (ii) speed

at maximum torque. Solution:

5.1=FL

st

TT

and 2max =FLT

T then, 75.0

25.1

max==

TTst

75.012

2max max

max =+

=T

Tst

ss

TT

Then, 075.0275.0maxmax

2 =+− TT ss

Then 21525.2max

=Ts (unacceptable) Or 451416.0max

=Ts

2*2

max

max

22max =

+=

FLT

FLT

FL ssss

TT

But 451416.0max

=Ts

Then 2*451416.0*2

451416.0 22max =

+=

FL

FL

FL ss

TT

0451416.0451416.0*4 22 =+− FLFL ss

0203777.080566.12 =+− FLFL ss

6847.1=FLs (unacceptable) or 120957.0=FLs

rpmns 15004

50*120==

then (a) ( ) sFLFL nsn *1−=

( ) rpmnFL 13191500*120957.01 =−=

(b) ( ) ( ) rpmnsn sTT 8231500*451416.01*1maxmax

=−=−=

Example 4.10 A 3φ, 280 V, 60 Hz, 20 hp, four-pole induction

motor has the following equivalent circuit parameters.

12.01 =R Ω, 1.02 =′R Ω, 25.021 =′= XX Ω, and 10=mX Ω

The rotational loss is 400 W. For 5% slip, determine (a) The

motor speed in rpm and radians per sec. (b) The motor current. (c)

The stator cu-loss. (d) The air gap power. (e) The rotor cu-loss. (f)

The shaft power. (g) The developed torque and the shaft torque.

(h) The efficiency.

rpmns 18004

60*120== , sec/5.1882*

601800 rads == πω

Solution:

0.12 j0.25

j10

j0.25

205.01.0

=

Ω Ω

Ω

Ω

ee XRjZ +++= 25.012.01

( ) o

jjjjZ 55.231314.225.102

25.02*1025.012.01 ∠=+

+++= Ω

1.1203

2081 ==V V

ooI 55.231314.2

55.231314.21.120

1 −∠=∠

= A

(c) WP 031.114312.0*3479.56*3 21 ==

(d) ( ) WP os 9794.1861055.23cos*3479.56*1.120*3 =−=

WPPP sag 9485.174671 =−=

(e) WsPP ag 3974.8739785.17467*05.02 ===

(f) ( ) WPsP agm 5511.165941 =−=

(g) mNP

T ag .6682.925.188

9485.174675.188

===

NmP

T shaftshaft 9127.85

5.1885511.16194

5.188===

(h) %02.87100* ==s

shaft

PP

η

Example 4.11 A 30, 100 WA, 460 V, 60 Hz, eight-pole induction

machine has the following

equivalent circuit parameters:

07.01 =R Ω, 05.02 =′R Ω, 2.021 =′= XX Ω, and 5.6=mX Ω

(a) Derive the Thevenin equivalent circuit for the

induction machine.

(b) If the machine is connected to a 30, 460 V, 60 Hz supply,

determine the starting torque, the maximum torque the machine

can develop, and the speed at which the maximum torque is

developed.

(c) If the maximum torque is to occur at start, determine the

external resistance required in each rotor phase. Assume a

turns ratio (stator to rotor) of 1.2.

Solution: VVXX

XVm

mth 7.2576.265*

5.62.05.6* 1

1=

+=

+=

( ) ( ) 1947.006589.05.62.007.0

07.02.0*5.6 jjj

jjjXR thth +=+++

=+ Ω

0.06589 j0.1947 j0.2

s05.0

Ω Ω Ω

257.7V

(b) ( ) ( )[ ]

NmTst 7.6242.01947.005.006589.025.94

05.0*7.257*322

2

=+++

=

( )[ ]Nm

T

8.22672.01947.006589.006589.025.94*2

7.257*322

2

max

=

+++=

( )1249.0

2.01947.006589.0

05.022max

=++

=Ts

Speed in rpm for which max torque occurs

= ( ) ( ) rpmns sT 5.787900*1249.01*1max

=−=−

(c) ( )

2221

21

2max

RXXR

RsT ′′++

′= α

or 4.005.0*1249.01

*1

22max

==′=

=′ Rs

sR

T

startstart Ω

Then ( ) 243.02.1/05.04.0 2 =−=extR Ω

UNIT-4Starting & Speed control of 3ph Induction Motor

Presented byC.GOKUL

AP/EEE

UNIT-4 Syllabus

Necessity of Starters / NEED FOR STARTING

Why we need starters? As it is seen that a 3 phase induction motor has

positive finite starting torque ‘T’ when slip s=1. thismean that 3-pahse induction motor is a self-startingmotor and begins to rotate on its own whenconnected to a 3-phase supply.

At the instant of starting 3-phase induction motorbehaves like a transformer with a short-circuitedsecondary.

Consequently, a 3-pahse induction motor takes highstarting current if started at full voltage. In order tolimit this high starting current to reasonable limitsstarting methods are used.

STARTING METHODS OF INDUCTION

MACHINE

Methods of Starting There are primarily two methods of starting the

induction motor:-a) Full voltage starting.b) Reduced voltage starting. Full voltage starting methods consist of:-a) DOL (Direct-on-line starting) Reduced voltage starting consist of:-a) Stator resistor (or reactor) starting.b) Auto-transformer starting.c) Star-delta starting.

AUTO TRANSFORMER

STARTER

Fig: Auto-transformer starting

Stator

Rotor

The fraction of xV1 is applied to the stator wdg at starting.

1V

1xV

1. Voltage is changed by transformer action

2. So power loss and input current are less.

xV1

LIscst xII =

As speed increases, gradually voltage is increasedFinally full voltage is applied to the motor.Advantages

and not by dropping voltage as that of reactor


Stator

Rotor

The stator starting current is

1V

1xV

sc11st xIzxVI == /

xV1

LIscst xII =

For auto-transformer, input VA= output VAILV1=Ist (xV1)

IL=xIst

IL=x2Isc

Therefore, line current atinput is x2 times the DOLcurrent.

fl

2

1fl

1st sII

=

Test

Tefl fl

2

fl

sc2 sIIx

=Thus,


Stator

Rotor

Line current at input due to auto-transformer starting

1V

1xVxV1

LIscst xII =

Line current at input due to stator reactor starting =x

Stator

Rotor1V

1xVxV1

LIscst xII =


Stator

Rotor

Line current at input due to auto-transformer starting

1V

1xVxV1

LIscst xII =

Line current at input due to stator reactor starting =x

Starting torque with auto transformer startingStarting torque with DOL starting =x2

Starting torque with auto transformer startingStarting torque with stator reactor starting =1

STAR DELTA STARTER

Star-Delta starting

Stator

Rotor

For delta, 6 terminals are required.For star, 3 terminals of stator wdg are required.

TPDT

R Y B

2- Run

1- Start - Star

- Delta

Fig.: Star-Delta starting

Now make deltaConnection.

At starting TPDT to 1, wdg in star

Stator

Rotor

Motor rotates.Reduced voltage is applied to wdg = VL/√3

TPDT

R Y B

2- Run

1- Start - Star

- Delta


The starting current is1Lst.y zVI 3/=

L.yI= Starting Line current

Now TPDT to 2- DeltaLine voltage appliedto wdg. Motor runs at rated speed

Rotor

Stator

TPDT

R Y B

2- Run

1- Start - Star

- Delta






Now TPDT to 2- DeltaLine voltage appliedto wdg Motor runs at rated speed

At starting, if, wdg in deltaThe starting current is

1Lst.d zVI /= sc.dI=

st.dL.d II 3=

st.dst.y II3

1=∴

Starting line current with Y-Δ starter

Thus Ist.y in star is one third of that current in delta.

=Ist.y

√3 Ist.d=

13Starting line current with stator in Δ





Now TPDT to 2- DeltaLine voltage appliedto wdg Motor runs at rated speed

In case of auto-transformer, if turn ratio x = 1/√3Then starting line current and is starting torque are

This shows that

This shows that Tst.y in star is one third of starting torque in delta.

Star delta starting is equivalent to auto transformer

Starting torque with Y-ΔstartingStarting torque with stator in Δ =

(V1/√3)2

V12 =

13

reduced to one third of their values with delta.

if auto transformer turn ratio x=1/√3=0.58 or 58% tappingThis method is cheap, effective and used extensivelyUsed for tool drives, pumps, motor-generator set.Used up to rating of 3.3kV, After this voltage, m/c becomes expensive for delta winding

exceed 1.5 times full load current. The short circuit currenton normal voltage is 4.5 times the full load current and the

full load torque.

that the supply current during starting of IM does not

Solution

Example Determine the % tapping of the auto-transformer so

full load slip is 3%. Calculate the ratio of starting torque

Stator

Rotor1V

1xVxV1

LIscst xII =

IL=1.5IFLIL=1.5IFL

Isc=4.5IFL

IL/Isc=0.333

In auto-transformer IL/Isc=x2 x=0.577Hence % tapping is 57%

Stator

Rotor1V

1xVxV1

LIscst xII =

fl

2

1fl

1st sII

=

Test

Tefl fl

2

fl

sc2 sIIx

=Now

( ) 0.034.50.333 2=

0.202=

delta starting. The supply voltage is 400V, full load effn is82% and full load power factor is 0.85% (lag).

Neglect magnetizing current.

its full load current, the stator of which is arranged for star

Solution

Example The short circuit line current of a 6hp IM is 3.5 times

Calculate the line current at the instant of starting.

6hp IM,

Star-delta startingIsc=3.5IFL

Isc (line) =3.5 IFL

Voltage =400VηFL=82%, pf=0.85 (lag)

P=√3 VLILcosθ

0.8540031

0.827466I

××

×

=LIFL=

=9.26A (line current fordelta)

=5.34A (phase current for delta)

=18.73AIsc=3.5IFL=3.5x5.34

At the instant of starting, motor wdg is in starFor star, line current is equal to phase current.IL at the instant of start =18.73A for delta (400V)IL at the instant of start =18.73/√3 A for star (400/√3)

=10.81A


DOL(Direct-on-line)Starter

DOL(Direct-on-line)starting This method involves direct switching of

poly-phase stator on to the supply mains. The motor takes starting current of 5 to 7 times its

full load current depending upon its size anddesign.

Such large current of short duration don’t harm therugged squirrel cage motor, but the high currentsmay cause objectionable voltage drop in powersupply feeding the induction motor

These large voltage drop causes undesirable dip inthe supply line voltage, consequently affecting theother equipments connected to the same supply.

The relation between the starting torque Ts and full load torque Tf is now obtained .

Let Is and If be the per phase stator currents drawn from the supply mains corresponding to starting and full load conditions respectively.

We know:-

Therefore:-

------Eqn(1)

Now

srIT

se

222 ..1

ω=

ff

s

ff

s

f

s sII

srIrI

TT .1

2

22

22

==

scsc

st IZVI == 1

V1 is per-phase stator voltage & Zsc=(r1+r2)+j(x1+x2), is the leakage impedance.

Therefore Eqn(1) can be written as:-

----Eqn(2)f

f

sc

f

s sII

TT .

2

=

Stator resistance(reactor)

Starter

Stator resistance(reactor)method In this method, a resistor or a reactor is inserted in

between motor terminals and supply mains. At the time of starting some voltage drop occurs

across the starting resistor and therefore only afraction ‘x’ of supply voltage appears across it.

This reduces the per phase starting currents Isdrawn by the motor from the supply mains.

As the motor speeds up, thereactor is cut out in stepsand finally short-circuitedwhen the motor speed isnear to synchronous speed.

Since the per phase voltage is reduced to ‘xV1’ the per phase starting current is:-

Now we know:-

Therefore we have:-

-----Eqn(1) Therefore:-

scsc

s xIZxVI == 1

srIT

s

222.1

ω=

ff

sc

f

s sIIx

TT

2

2

=

22

1

1

switchingdirect with torquestartingstartingreactor with torquestarting x

VxV

=

=

Rotor resistanceStarter

ROTOR RESISTANCE STARTER(only for slip ring induction motor)


• Increasing the rotor resistance, not only is the rotor (and hence stator) current reduced at starting, but at the same time, the starting torque is also increased due to improvement in power factor.

• The introduction of additional external resistance in the rotor circuit enables a slip-ring motor to develop a high starting torque with reasonably moderate starting current.

• Hence, such motors can be started under load. This additional resistance is for starting purpose only. It is gradually cut out as the motor comes up to speed.

Speed control of 3 phase Induction Motor

Speed Control of IM

• Given a load T–ω characteristic, the steady-state speed can be changed by altering the T–ω curve of the motor

501

fPPs πωω 42

==

( )

++

+

=2

2'

2'3

lrlsr

s

s

s

re

XXs

RR

VsRTω

Pole Changing

Varying line frequency

Varying voltage (amplitude)2

3

1

a) By changing the applied voltage:Torque equation of induction motor is

Rotor resistance R2 is constant and if slip s is small then sX2 is so small that it can be neglected. Therefore, T ∝ sE2

2 where E2 is rotor induced emf and E2 ∝ V& hence T ∝ V2, thus if supplied voltage is decreased, torque decreases and hence the speed decreases.

This method is the easiest & cheapest, still rarely used because-1) A large change in supply voltage is required for

relatively small change in speed.2) Large change in supply voltage will result in large

change in flux density, hence disturbing the magnetic conditions of the motor.

b) By changing the applied frequency• Synchronous speed of the rotating magnetic field of

induction motor is given by,

f = frequency & P = number of stator poles.• Thus, synchronous speed changes with change in

supply frequency, and thus running speed also changes. However, this method is not widely used. This method is used where, only the induction motor is supplied by a generator (so that frequency can be easily change by changing the speed of prime mover).

V/F control

Variable Frequency Control of IM (v/f control) Speed control above rated (base) speed

Requires the use of PWM inverters to control frequency of motor Frequency increased (i.e. ωs increased) Stator voltage held constant at rated value Air gap flux and rotor current decreases Developed torque

decreasesTe ∝ (1/ωs)

For control belowbase speed –use Constant Volts/Hz method

506

Constant Volts/Hz (V/f) Control Airgap flux in the motor is related to the induced stator

voltage E1 :

For below base speed operation: Frequency reduced at rated Vs - airgap flux saturates

(f ↓ ,φag ↑ and enters saturation region oh B-H curve):- excessive stator currents flow- distortion of flux wave- increase in core losses and stator copper loss

Hence, keep φag = rated flux stator voltage Vs must be reduced proportional to reduction

in f (i.e. maintaining Vs / f ratio)

507

fE

ag1=φ

fVs≈ Assuming small voltage drop

across Rs and Lls

Constant Volts/Hz (V/f) Control Max. torque remains almost

constant For low speed operation:

can’t ignore voltage drop across Rs and Lls (i.e. E1 ≠ Vs)

poor torque capability(i.e. torque decreased at low speeds shown by dotted lines)

stator voltage must be boosted – to compensate for voltage drop at Rs and Lls and maintain constant φag

For above base speed operation (f > frated): stator voltage maintained at

rated value Same as Variable Frequency

control (refer to slide 13)508

s

sVTω

2

max ∝f

Eag

1=φf

Vs≠

Constant Volts/Hz (V/f) Control

509

Vrated

frated

Linear offset

Non-linear offset – varies with IsBoost

Vs vs. f relation in Constant Volts/Hz drivesVs

f

Linear offset curve –• for high-starting

torque loads• employed for most

applications

Non-linear offset curve –• for low-starting

torque loads

Boost - to compensate for voltage drop at Rsand Lls


• For operation at frequency K times rated frequency:– fs = Kfs,rated ⇒ ωs = Kωs,rated (1)

(Note: in (1) , speed is given as mechanical speed)

– Stator voltage: (2)

–Voltage-to-frequency ratio = d = constant:

(3)

510

rated,

rated,

s

sVd

ω=

><

=rated,rated,

rated,rated,

when , when ,

sss

ssss ffV

ffKVV


For operation at frequency K times rated frequency:

Hence, the torque produced by the motor:

(4)

where ωs and Vs are calculated from (1) and (2) respectively.

511

( )

++

+

=22

2'

2'3

lrlsr

s

s

s

re

XXKs

RR

VsRTω


For operation at frequency K times rated frequency:The slip for maximum torque is:

(5)

The maximum torque is then given by:

(6)

where ωs and Vs are calculated from (1) and (2) respectively.

512

( )222

'

max

lrlss

r

XXKR

Rs++

±=

( )

++±

=222

2

max 23

lrlsss

s

s XXKRR

VTω


513

Field Weakening Mode (f > frated)• Reduced flux (since Vs is constant)• Torque reduces

⇒Constant Power Area(above base speed)

Constant Torque Area

(below base speed)Rated (Base) frequency

Note: Operation restricted between synchronous speed and Tmax for motoring and braking regions, i.e. in the linear region of the torque-speed curve.


514

Constant Power Area

Constant Torque Area

c) By changing No. of poles

synchronous speed(Ns) (and hence, running speed) can be changed by changing the number of stator poles. This method is generally used for squirrel cage induction motors, as squirrel cage rotor adapts itself for any number of stator poles. Change in stator poles is achieved by two or more independent stator windings wound for different number of poles in same slots.For example, a stator is wound with two 3phase windings, one for 4 poles and other for 6 poles.For supply frequency of 50 Hzi) synchronous speed when 4 pole winding is connected,

Ns = 120*50/4 = 1500 RPMii) synchronous speed when 6 pole winding is connected,

Ns = 120*50/6 = 1000 RPM


CASCADING OPERATION

Cascaded connection

• In this method of speed control, two motors are used. Both are mounted on a same shaft so that both run at same speed.

• One motor is fed from a 3phase supply and other motor is fed from the induced emf in first motor via slip-rings.

Motor A is called main motor and motor B is called auxiliary motor.Let, Ns1 = frequency of motor A

Ns2 = frequency of motor BP1 = number of poles stator of motor AP2 = number of stator poles of motor BN = speed of the set and same for both motorsf = frequency of the supply

Now, slip of motor A, S1 = (Ns1 - N) / Ns1.frequency of the rotor induced emf in motor A, f1 = S1f

now, auxiliary motor B is supplied with the rotor induce emftherefore, Ns2 = (120f1) / P2 = (120S1f) / P2.

now putting the value of S1 = (Ns1 - N) / Ns1

• At no load, speed of the auxiliary rotor is almost same as its synchronous speed.i.e. N = Ns2.

Four different speeds can be obtained1. when only motor A works, corresponding

speed = Ns1 = 120f / P12. when only motor B works, corresponding

speed = Ns2 = 120f / P23. if cummulative cascading is done,

speed of the set = N = 120f / (P1 + P2)4. if differential cascading is done,

speed of the set = N = 120f (P1 - P2)

Slip power recovery

•Kramer•Scherbius

1) Kramer System

R Y B

MIM

f

ACM

Voltage regulating device

If brush emf is more than slip voltagePower flows from ACM-Rotor of MIM.MIM operates at Super-Synchronous speedIf brush emf is less than slip voltagePower flows from Rotor of MIM- ACM.MIM operates at Sub-Synchronous speedSince power is flowing from one machine to another with one shaft, it is constant power drive.

2) Scherbius System

Power changes

At Super-Synchronous speed, power flows from supply-AIM (Motor) - ACM -rotor of MIM.

R Y B

MIM

R Y B

ACM AIM

Voltage regulating device

Constant torque drive

At Sub-Synchronous speed, power flows from rotor of MIM - ACM – AIM (Gen) - supply.

f

Braking of 3ph Induction

Motors•Plugging•Dynamic Braking•Regenerative Braking

Braking Methods

• Regenerative Braking• Plugging or reverse voltage braking• Dynamic ( or rheostatic ) braking :

a) ac dynamic brakingb) Self-excited braking using capacitorc) dc dynamic brakingd) zero-sequence braking

1. Regenerative Braking• If an induction motor is forced to run at speeds in

excess of the synchronous speed, the load torque exceeds the machine torque and the slip is negative, reversing the rotor induced EMF and rotor current. In this situation the machine will act as a generator with energy being returned to the supply.

• If the AC supply voltage to the stator excitation is simply removed, no generation is possible because there can be no induced current in the rotor.

Regenerative braking• In traction applications, regenerative braking is

not possible below synchronous speed in a machine fed with a fixed frequency supply. If however the motor is fed by a variable frequency inverter then regenerative braking is possible by reducing the supply frequency so that the synchronous speed becomes less than the motor speed.

• AC motors can be microprocessor controlled to a fine degree and can regenerate current down to almost a stop

1 0Ns0

Te 4 poles8 poles

TLSpeedSlip

A

B

C

D

+Te-Te

TL

Two quadrant operation

A

B

D

Regenerative braking• Power input to induction motor:Pin=3VIscosφs

Motoring operation φs<90º ωm< ωms

Braking φs>90º ωm> ωms

Regenerative braking

• Advantage: Generated power is usefully employed

• Disadvantage: It can not be employed below synchronous speed when fed from constant frequency source.

• Speed Range : Between synchronous speed and the speed for which braking torque is maximum.

2. Plugging• Plugging induction motor braking is done by reversing

the phase sequence of the motor. Plugging braking of induction motor is done by interchanging connections of any two phases of stator with respect of supply terminals. And with that the operation of motoring shifts to plugging braking.

• During plugging the slip is (2 - s), if the original slip of the running motor is s, then it can be shown in the following way.


• From the figure beside we can see thatthe torque is not zero at zero speed.That’s why when the motor is needed tobe stopped, it should be disconnectedfrom the supply at near zero speed.

• The motor is connected to rotate in thereverse direction and the torque is notzero at zero or any other speed, and as aresult the motor first decelerates to zeroand then smoothly accelerates in theopposite direction.

3. DC Dynamic Braking

The disadvantages of plugging are removed in dynamic braking.

or Rheostatic or AC Dynamic Braking

Under normal operating conditionDynamic braking requires less power.

Stator -

Rotor -

Rotating Magnetic Field - Ns

Rotates - NrTe -

If DC supply is given to statorStator -

Rotor -

Stationary Magnetic Field -Ns =0

Rotates -Nr

Faster sNs

Slower

Faster Ns(1-s)=

Slower

NsSThis Teb is dynamic braking torque.Teb depends on 1. DC source. 2. Rotor resistance 3. Speed

Teb

Consider rotor is running at syn speed NsStator is excited by DCThe relative speed between stator field and rotor is Ns.Slip = (Relative speed Ns)/Ns = 1

This is equivalent to IM with a rotor at STANDSTILLNow consider, rotor is at rest and stator is excited by DCStationary flux induces no rotor emfThis is equivalent to IM with a rotor RUNNING at Syn speedConclusions1. Rotor at syn speed with DC dynamic braking is similar to

rotor at rest during IM operation2. Rotor at rest with DC dynamic braking is similar to

rotor running at syn speed during IM operation

Circuit DiagramDC

Rotor

Stator

AC

R1

R1 is connected to limit stator current

Additional rotor resistance is also connected to limitthe current and to obtainbraking characteristics

Circuit Diagram

Stator

AC

Rotor

Rectifier

TransformerR1

Under normal operating conditionRotor speed w r t stator field under DC dynamic braking is Ns(1-s) = NsS

In the equivalent ckt diagram, replace s by S

r2/S

jx2

jXm

I1 I2

IΦ

VDC

x1 = 0, and no stator core loss

V1

SE2

I0

I2

I1I1’

I2r2

jI2Sx2

In phasor diadramalso replace s by S

The dynamic braking torque isTed =

The T-s ch is similar to IM but with slip scale reversed

SrI3 22

2sω

1 0Ns0

Te

TLSpeedSlip

A

r2R2’R2’’R2’’’

R2’ < R2’’< R2’’’

Ted increases with increase in rotor circuit resistance

Ted

Due to this it is also called as RHEOSTATIC braking

The entire power developed in rotor is dissipated in R2

MMF produced by 3-ph wdg due to AC

MMFAC = NI23

m

MMF produced by single ph due to DC = NIDC

The resultant MMF produced due to DC

IDC NIDC N

60

√3IDC N

MMFDC = NIDC3

DCandACtodueMMFequalFor∴NINI

23

DCm 3=

NI23I 1DC =

AC dynamic braking in nothing but SEIG operationA bank of capacitors is connected across three phases of stator wdg.

IG receives AC excitation from bank of capacitor

AC

CC

CStator

Rotor

The generated electrical energy isdissipated as heat in rotor circuitDue to high cost of capacitor, thismethod is not used in practice.

Advantages of Dynamic Braking1. Smooth stop2. Less rotor ckt

loss3. No tendency to

reverseDisadvantage: Less quick than plugging

UNIT-5Single phase Induction

Motor & Special Machines

Presented byC.GOKUL

AP/EEE

Single phase Induction Motor


Introduction

• What is single phase induction motors?is an induction motor having a squirrel cage

rotor and a single phase stator winding.

Working Principle

Suppose the rotor is at rest and a single phase supply is given to the stator winding. Now the current flowing in the stator winding will produce a m.m.f with in the stator and this m.m.f induces a current in the rotor. Again the induced current inside the rotor will produce a m.m.f with in the rotor itself which is equal in magnitude and opposite in direction with the stator m.m.f. Thus the two m.m.f cancel out each other and as a result there will be no net torque acting on the rotor. There for the rotor will stay at rest. So due to this effect, we have to find another method to start the motor.

Types of Single Phase Induction Motors

Depending on the method used to start the motor : -

• 1) Capacitor-start motors• 2) Capacitor-run motors• 3) Capacitor start-and run motors • 4) Shaded-pole motors

1) Capacitor-Start phase induction motor

• A capacitor-Start motor is a spilt phase induction motor with a starting capacitor inserted in series with the start winding creating an LC Circuit which is capable of producing a much greater torque.

• An Lc circuit refers to a circuit containing an inductor w/c connected together they can act as an electrical resonator w/c stores electrical energy.

Working Principle of single phase capacitor-start motor

• In capacitor-start motors the capacitor enables the motor to handle heavy start loads by increasing the strength of the magnetic field created by the windings. The capacitor is individually mounted outside of the motor as a separate unit either on the top or side of the motor with a centrifugal switch located between the capacitor and the start winding. The switch connects the capacitor with the motor at startup and disconnects them when the motor has reached about 75% of its operating speed. And during startup period when the centrifugal switch is closed, capacitor-start motors typically deliver from 250-350% of the full load torque.

Equivalent circuit of capacitor-start motors

Types of single phase capacitor-start motor

Among this the basic types include:-A) Single voltage externally reversible B) Single voltage non reversible

Applications of single phase capacitor-start motors

Capacitor-start from high torque (>175% full load) are used: - Operation having high starting loads

such as: - Elevators- Compressors &- Refrigerators

• Capacitor-start moderate torque (<175% full load) are used:- Operation having low starting

loads such as:- Fans- Blowers &- Small pumps.

2) Capacitor run motorsCapacitor-run motors are motors having a

capacitor connected in series with the start winding in order to increase the running efficiency.

Capacitor-run motors use run-capacitors that are designed for continuous duty which are energized the entire time during operation of the motor.

Working principle of single phase capacitor-run motors

• In capacitor-run motors, a run-capacitor is connected to the start winding of the motor and it constantly energizes the start winding while the motor is running. And this creates a 90o phase change between the start winding current and the run winding current making a two phase motor. As a result a rotating magnetic field is created within the motor which causes the rotor to rotate more efficiently.

Advantages and disadvantages of capacitor-run motors

AdvantagesThe capacitor remains in the circuit at all

times thus no centrifugal switch is required.They can be designed to have low vibration

and less noise under full load condition.If properly designed, they are more efficient

than other type of motors.

DisadvantagesSince capacitor start motors have low

starting torque they cannot be used in applications with severe starting conditions.

Application of single phase capacitor-run motors

are mainly used for applications requiring low starting torque and high efficiency such as:- Small compressors,

Pumps & Fans.

3) Capacitor start-and-run motors

Capacitor-start-and-run motors or permanent-split capacitor motors are single phase induction motors having

capacitors connected in the circuit during both the starting and the running period. In this type of motors both the start winding and the run windings are permanently connected to the power source through a capacitor at all times.

Types of single phase capacitor start-and-run motors

Depending on the number of capacitors used: -1. Single value capacitor start-and-run

motors:

2. Two value capacitor start-and-run motors

The two values of capacitance can be obtained using two different methods.

a. By using two capacitors in parallel b. By using a step up transformer

Advantage

• Ability to start heavy loads• Ability to develop 25% overload capacity• Higher efficiency and power factor • Extremely quiet operation

Applications of single phase capacitor start-and-run motors

• Two value capacitor start and run motors are frequently used in applications requiring variable speed such as : -

Air handlers, Blowers and Fanes.

• Single value capacitor start-and-run motors are used in applications requiring low starting torque such as:-

Fans Blowers & Voltage regulators.

4) Shaded pole motors

• A shaded pole motor is a single phase induction motor having one or more short circuited windings acting only on a portion of the magnetic circuit.

• Generally the winding is a closed copper ring embedded in the face of the pole together known as the shaded pole which provides the required rotating field for starting purpose.

Working principles of single phase shaded pole motors

• Now when an alternating current is passed through the field or main winding surrounding the whole pole, the magnetic axis of the pole shifts from the unshaded part to the shaded part. which is analogous with the actual physical movement of the pole. As a result the rotor starts rotating in the direction of this shift from the unshaded part to the shaded part.

Advantages and disadvantages of single phase shaded pole motors

AdvantagesSimple in construction Tough surface Reliable and cheap

DisadvantagesLow starting torque Very little overload capacityLow efficiency (5% for tiny sizes – 35%

for higher ratings)

Applications of single phase shaded-pole motors

• Because of its low starting torque, the shaded pole motor is generally used for

Small fans,Toys, Hairdryers,Ventilators etc.

Special Machines

There are variety of special machines availableHere, our territory includes Stepper Motor Hysteresis Motor AC series Motor Linear Reluctance Motor Repulsion Motor

Stepper Motor

Stepper Motor

Stepper Motor, derives its name from the fact that it follows definitive step in response to input pulses

See to it, that the input is in the form of pulses Straightaway it is understood that the input, being pulses, can

be controlled and in turn the output gets controlled Wherever precise positioning is required stepper motors are

widely employed Typical values – stepper motors develop torque ranging from 1

µN-m upto 40 µN-m – power output range from 1 W to 2500 W

Stepper Motor – types

There are three designs of stepper motors available in the literature

They are Variable Reluctance stepper motor Permanent magnet stepper motor Hybrid stepper motor

Stepper Motor – Variable Reluctance Stepper Motor

Operating principle

1. Variable Reluctance Stepper Motor As usual, it has Stator Rotor

Variable Reluctance Stepper Motor- Stator

Stator is a hollow cylinder whose inner periphery houses salient poles

Variable Reluctance Stepper Motor- Rotor

Rotor is a solid cylinder whose outer periphery has salient poles

Variable Reluctance Stepper Motor


When we emphasize that the operation just performed is 1-phase-ON mode we indirectly mean that we have something called as 2-phase-ON mode and so on

As the name goes, 2-phase-ON mode denotes 2 phases being switched ON at the same time


2-phase-ON mode



When we started, 2-phase-ON mode, many would have thought that step angle would be 15 deg

But the table in the previous slide shows the step angle is same as that of the previous case (30 deg, maintained)

But the position of the rotor is changed, which is a desirable factor in some of the position control experiments


If the step angle is altered from the original intended design, it would add much to the application side of our machine

Can we bring any other step size here? Is it possible, first of all? The answer is yes, it is possible There is no restriction imposed on us in altering the

combination of switching pulses In fact, the 2-phase-ON mode is the child of our manipulation of

combination of phases involved in switching


Going by the same discussion, if we resort to the combination of 1-phase-ON mode and 2-phase-ON mode we will end up with some interesting operation



It is interesting to note here that this discussion has no end in it

We have something called as micro-stepping and the reader is advised to do it as an assignment


All the previous slides regarding Variable Reluctance Stepper Motor can be confined to what is referred to as single-stack variable reluctance stepper motor

It becomes clear by now that we have something called as multi-stack variable reluctance stepper motor


2. Permanent Magnet Stepper Motor

It is very similar to Variable Reluctance stepper motor

The only difference being that the rotor is made up of permanent magnet

In VR motors, the rotor is a magnetic material (It can carry the flux lines through it)

Permanent Magnet Stepper Motor

Stator and Rotor


The stator phases can be excited with either positive current or negative current

Positive current in phase A will create a set of poles while the negative current will create opposite poles

Similar is the case with phase B


Consider positive current in phase A



Advantages Permanent magnets require no external

exciting current – low power loss High inertia Develops more torque than VR motor


Disadvantages It is very difficult to produce permanent

magnet rotor with more number of poles This makes the design of PM motors with

higher step angle

3. Hybrid Stepper Motor

Hybrid stepper motor combine the features of VR and PM stepper motors

The stator is an electromagnet The rotor is a permanent magnet The difference in the rotor is that the rotor

magnet is axial with one end completely north pole and other, south pole

Hybrid Stepper Motor

The confusion, if any, can be better illustrated with the schematic representation given below



The side view of the axial permanent magnet in the rotor is shown below


The operation is left as an assignment for the reader

The reader can build on this idea that the rotor alignment would be based on the attraction between formed stator poles (this being electromagnet) and permanent rotor poles

Applications

Type-writers Tape drives Floppy disk drivers Process control systems X-Y plotters

Hysteresis Motor

Hysteresis Motor This is based on the principle of hysteresis Basically this is a constant speed motor

similar to synchronous motor As is always the caseIt has a Stator and a Rotor

Hysteresis Motor - Stator

Stator of hysteresis motor is similar to single phase induction motor

The stator winding can be either split phase type of shaded pole type

Hysteresis Motor - Rotor

Rotor is a smooth cylinder made of chrome-steel

Materials of high resistivity comparable to that of an insulator are normally chosen to make the eddy current loss zero which make the core loss equal only to hysteresis loss

Hysteresis Motor – Operating Principle

The concept of hysteresis is the basis of such motor

As we know, hysteresis is the lagging of magnetic flux density (B) with respect to magnetizing field strength (H)

Hysteresis loss

Remember, the very old hysteresis curve It looks as shown

Hysteresis loss

At the outset, what we can elaborate is that if an attempt is made to induce pole in a magnetic material with higher retentivity – the induced pole will not loose its magnetic property even though the induction is taken out completely

It is like remembering some event even after the event is over (retaining something)

Hysteresis loss

Where this come in the machine?

The typical hysteresis loop for the rotor material

Torque – Slip characteristics

The torque – slip characteristics of hysteresis motor has some interesting points to note

Applications

Precision Audio equipments Record players Electric clocks Tele-printers

AC series Motor

AC series motor An AC motor with commutator and brush

assembly is referred to as commutator motor (Remember commutator and brush assembly

in the wound rotor of an Induction motor) There are two types of commutator motor AC series motor Universal motor

AC series motor

What would happen if a normal dc series motor is connected to ac supply?

The motor will run normally as the torque will still be unidirectional

This is due to the fact that current and flux will change direction simultaneously (dealt during 3 ph IM)

But, power factor would be very poor due to very high inductance of armature and field windings

At the same time, alternating flux would induce eddy emf in the core leading to heavy eddy current loss in the machine

Also, sparking occur at brushes during the commutation period due to heavy voltage and current

AC series motor

These disadvantages make the machine unsuitable with AC supply

Proper modifications can make the machine suitable with AC supply

AC series motor – Required Modifications

Eddy current loss and the associated heating loss can be overcome by properly laminating the machine’s armature core and field core

The power factor can be controlled by decreasing the reactance of armature winding and field winding

AC series motor – Modifications Elaborated

Decreasing the reactance of the field winding increases the speed of the machine due to reduction in the air gap flux

Increase in the speed gives rise to decreased torque Now to improve the torque, armature turns has to be

increased proportionately But this will again increase the effective reactance of

the machine which is undesirable

AC series motor – Modifications Elaborated

To keep the armature reactance minimum and the associated armature reaction reactance effect, a special compensating winding is provided

The compensating winding is connected in such a way so that the flux produced by the compensating winding will be exactly in opposition to the flux produced by the armature winding

This will neutralize the armature reaction reactance effect

AC series motor – Modifications Implementation

This compensating winding can be connected in two ways Based on the connection it is referred as conductively

compensated and inductively compensated


Another major set back is the sparking associated with commutation

In dc motors, this is overcome by commutating poles (com poles) or inter poles

The voltage induced in the short circuited armature winding is huge enough (this voltage is absent in the case of dc motors) which creates undesirable sparking even when inter poles are provided


One method to reduce sparking connecting a shunt resistance with the commutating winding of the machine

By adjusting the resistance, voltage across the compole winding is adjusted

AC series motor – Characteristics

The characteristics of AC series motor are very similar to dc series motor

Repulsion Motor

Repulsion Motor It has a Stator Rotor

Repulsion Motor - Stator

Stator is a hollow cylinder whose inner periphery houses armature conductors

Winding is excited with single phase supply

Repulsion Motor - Motor

Rotor is a solid cylinder whose outer periphery has conductors

It is very similar to the armature of the dc motor with commutator and brush arrangement

The brushes are short circuited by low resistance jumper (why?)

Repulsion Motor - Operation

The operation of the repulsion motor is shown with stator designed as salient pole type

The operation will remain same with stator discussed as salient pole type

But take it that the stator is distributed type with slots carrying single phase armature conductors


To make it clear


Then, how to make the motor start?

Repulsion Motor - Shortcomings

Speed changes as the load is changed It becomes very high (dangerously high) at no load Working power factor is very poor Likely sparking at brushes

Repulsion Motor - Shortcomings

Speed changes as the load is changed It becomes very high (dangerously high) at no load Working power factor is very poor Likely sparking at brushes


Repulsion Motor – Overcoming the disadvantages

An attempt in overcoming the disadvantages has given way to new types of Repulsion motor

Compensated Repulsion Motor Here, an extra winding called the compensating

winding is added in series with the armature winding This winding is placed in the inner slots of the stator The main purpose of compensating winding is to

improved the power factor (as in the case of compensation provided in the AC series motor) and to improve the speed regulation


Repulsion-start Induction-Run Motor As the name indicates the motor starts as a

repulsion motor and after attaining 75 percent of the speed the brushes are lifted and the armature winding is shorted as Induction Motor

This arrangement is advantageous as the brushes would not any current during operation

There are also designs in which the brushes ride on the commutator throughout the operation


Repulsion Induction Motor This is the third design in which stator is the same as

in normal repulsion motor But the rotor has two separate windings One winding carries commutator and brush

arrangement similar to dc motor Other winding is squirrel cage winding similar to

cage induction motor Both these windings operate during the entire period

of operation of the motor


Squirrel cage windings are placed deep inside the rotor and remains inactive during start due to its high reactance

When the rotor attains 85 % of the speed, squirrel cage windings takes control

Commutated windings provide the starting torque which is seen to well above 350 percent of the full-load torque

Linear Induction Motor

Linear Induction Motor The readers are advised to do this part as an

assignment Interested people can this important points before

taking up the assignment



A normal Induction motor has a stator and a rotor Stator is a hollow cylinder with conductors in its inner

periphery Rotor is a solid cylinder with conductors on its outer

periphery


If the stator is cut in to half parallel to its axis (It will look as english alphabet “U” from the front end), the motor is referred to as sector Induction Motor

The important to note is that the motor will work developing almost 30 % of its power rating

Anyway the voltage has to be reduced to prevent saturation since the number of conductors has been reduced to half of its original value


If the U shaped stator and cylindrical rotor is made flat, then the machine is referred to as Linear Induction Motor

As a passing reference, the reader can note that this type of machine is employed in trains which operate on the principle of Magnetic Levitation

Servo Motor


Introduction

They are also called control motors and have high-torque capabilities

Basic principle of operation is the same as that of other electromagnetic motors. However, their construction, design and mode of operation are different.

Their power ratings vary from a fraction of a watt up to a few 100 W.

Both DC and AC (2-phase and 3-phase) servomotors are used.Applications In radar , tracking and guidance systems, process controllers,

computers and machine tools.

DC Servomotors These motors are either separately-excited dc motors or permanent-

magnet dc motors. The schematic diagram of a separately-excited DC motor along with its

armature and field MMFs and torque/speed characteristics is shown in Fig. 39.26. The speed of DC servomotors is normally controlled by varying the armature voltage. Their armature is deliberately designed to have large resistance so that torque-speed characteristics are linear and have a large negative slope as shown in Fig. 39.26 (c). The negative slope serves the purpose of providing the viscous damping for the servo drive system.

As shown in Fig. 39.26 (b), the armature mmf. and excitation field mmf are in quadrature. This fact provides a fast torque response because torque and flux become decoupled.

Accordingly, a step change in the armature voltage or current produces a quick change in the position or speed of the rotor.

AC Servomotors Such motors normally run on a frequency of 60 Hz or 400

Hz (for airborne systems). The stator has two distributed windings which are displaced from each other by 90º (electrical).

The main winding (also called the reference or fixed phase) is supplied from a constant voltage source, Vm∠ 0º (Fig. 39.27). The other winding (also called the control phase) is supplied with a variable voltage of the same frequency as the reference phase but is phase-displaced by 90º (electrical).

The control phase voltage is controlled by an electronic controller. The speed and torque of the rotor are controlled by the phase difference between the main and control windings. Reversing the phase difference from leading to lagging (or vice-versa) reverses the motor direction.

Magnetic Levitation

System- Introduction

Introduction

• What are Magnetic levitation systems?

Maglev. are devices that suspend ferromagnetic materials with the aid of electromagnetism. It has wide number of applications such as high-speed trains, aerospace shuttles, magnetic bearings and high-precision platforms.

System Block Diagram

Intel micrcontroller+

+

-

Set point

Referenceinput

E*(s)

DigitalController

InterfaceCircuitE(s)

Tszoh

InterfaceCircuit

MagneticLevitation

SystemU(s)

Actual Ball position Y(s)

Maglev Front Panel

References

• Electrical Machines-II by S. B. Sivasubramaniyan -MSEC, Chennai

• http://yourelectrichome.blogspot.in/• http://www.electricaleasy.com/p/electrical-

machines.html• www.scribd.com• www.slideshare.net

http://yourelectrichome.blogspot.in/



http://www.scribd.com/

http://www.slideshare.net/

References• Armature Reaction of Alternator by N.Karthikeyan

• BEE2123 ELECTRICAL MACHINES Muhamad Zahim• EE20A - Electromechanical Energy Conversion• Alternators and Synchronous Motors by Amit Mishra• Electrical Machines www.utm.my• INDUCTION MOTOR by MUHAMMAD WAQAR• Single phase Induction Motor

Magnetic Levitation by Tori Johnson and Jenna Wilson

http://www.utm.my/

Books Reference

• Electric Machinery by A.E. Fitzgerald Charles Kingsley, Jr.Stephen D. Umans

• Electrical Machines by Nagrath & Kothari• Electrical Machines by P.S.Bimbhra• Electrical Machines-II by Godse • Electrical Machines-II by Gnanavadivel