Electrical Machines 2 AC Machines

ELECTRICAL MACHINES – II (AC MACHINES) Presented by C.GOKUL AP/EEE Velalar College of Engg & Tech,Erode EMAIL: [email protected]

Syllabus EE6502 Electrical Machines -II

BOOKS Reference

LOCAL AUTHORS: {For THEORY use this books} 1.Electrical Machines-II by “Gnanavadivel” – Anuradha Publication 2. 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 

NPTEL Reference • Electrical Machines II by Dr. Krishna Vasudevan & Prof. G. Sridhara Rao Department 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 which can 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

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. 

dψ dϕ e= − = −N dt dt

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 material as 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

V= (t ) ⇒ i (t )=

Q C = V q (t ) 1 = C C dq (t ) = dt

∫ i (t ) dt

dv (t ) C dt

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

=

V L

3V ph

I L = I ph

Delta Connection

V L

= V ph

IL =

3I ph

Maxwell's Right Hand Grip Rule

Right Handed Cork Screw Rule

Generators

Input



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).

Mechanical

Magnetic

Electrical

Output

Electric Generator Mechanical Energy

G

Stationary magnets - rotating magnets - electromagnets

Electrical Energy

Motor 

The Motor converts electrical power into mechanical power.

Electrical Energy

Input

M

Magnetic Electrical

Mechanical Energy

Mechanical

Output

Basic Construction Parts Stationary Part

Stator

Armature

Electrical

Mechanical

Rotor Rotating Part

Field

AC MACHINES Two categories: 1.Synchronous Machines: 



Synchronous Generators(Alternator)

Primary Source of Electrical Energy  Synchronous Motor 

2.Asynchronous Machines(Induction Machines)

UNIT-1 Synchronous 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 winding Rf= 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

a. Salient-Pole 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 d-axis

Non-uniform air-gap

N

D ≈ 10 m q-axis

S

S

Turbine Hydro (water)

Hydrogenerator

N

• 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

b. Cylindrical-Rotor(Non-Salient Pole)

D≈1m

Turbine

L ≈ 10 m Steam

d-axis Stator winding



High speed



3600 r/min ⇒ 2-pole



Uniform airgap Stato r

1800 r/min ⇒ 4-pole

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

N

q-axis

Rotor winding Roto r

Rating up to 2000 MVA S

Turbogenerator

• 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}

Frequency of Induced EMF 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

p n pn f = . = 2 60 120 N=Rotor speed in r.p.m P=number of rotor poles f=frequency of induced EMF in Hz No of cycles/revolution = No of pairs of poles = P/2 No of revolutions/second = N/60 No 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. Presented by C.GOKUL,AP/EEE Velalar College of Engg & Tech , Erode

Winding Factors( K , Kd) p

K

p

= cos

α

2  m  sin   2     m sin   2  

β

Kd

=

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

Vectorsum Kp = Arithmaticsum

Pitch factor – contd. 

Arithmatic sum

Pitch factor – contd. 

Vector sum

Pitch factor – contd.

Pitch factor – contd.

Kp

=

Vector _ sum = Arithmatic _ sum

2 Es cos 2 Es

= cos

α

2

α 2

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

emf _ with _ distributed _ winding Kd = emf _ with _ concentrated _ winding

β

0

180 180 = no.of _ slots _ per _ pole n

0



For calculating Vector sum

Kd

Kd

 mβ 2 r sin  2  =  β m 2 r sin   2  mβ sin  2  =  β m sin   2

     

     

Problem: Distribution factor /Breadth factor

EMF Equation of Alternator

Equation of Induced EMF

Average emf induced per conductor = dφ / dt Here, dφ = φP  If P is number of poles and flux / pole is φ Weber dt = time for N revolution = 60 / N second Therefore,  Average emf = dφ / dt = φP / (60 / N) 

=

ϕ NP 60

Equation of Induced EMF – contd. We know,  N = 120 f / P Substituting, 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 unity II.) When load p.f. is zero lagging III.) 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 compensate III.) 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 Ia ᴪ which denotes the angle between the Ef and Ia δ which denotes the angle between the Ef and Vt ra which denotes the armature per phase resistance Two 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, Xs The 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 voltage behaviour of two generators is by their voltage regulation (VR). The VR of a synchronous generator at a given load, power factor, and at rated speed is defined as

VR =

Enl − V fl V fl

× 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. Method Ampere-turns method / M.M.F. method ZPF(Zero Power Factor) Method / Potier ASA 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 capable of driving the alternator at its synchronous speed. The armature is connected to the terminals of a switch. The other terminals of the switch are short circuited through an ammeter. The voltmeter is connected across the lines to measure the open circuit voltage of the alternator.  The field winding is connected to a suitable d.c. supply with rheostat connected in series. The field excitation i.e. field current can be varied with the help of this rheostat. The circuit diagram is shown in 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 speed The 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 shortcircuit 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 field rheostat is brought to maximum position, reducing field current to a minimum value.  The T.P.S.T switch is closed. As ammeter has negligible resistance, the armature gets short circuited. Then the field excitation is gradually increased till full load current is obtained through armature winding.  This can be observed on the ammeter connected in the armature circuit. The graph of short circuit armature current against field current is plotted from the observation table of short circuit test. This graph is called short circuit characteristics, 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

Z s ,unsat = Ef or Vt (V)

Air-gap line OCC

Vrated

Isc (A) SCC

VA

IfB

2 s ,unsat

=

V A (= E f

)

I scA

X s ,unsat = Z s2,unsat − Ra2 : Ra is known from the DC test.

Isc,B

IfA

R +X 2 a

Isc, A If (A)

Since Xs,unsat>>Ra,

X s ,unsat ≈

Ef I scA

=

Vt , oc I scA

Xs under saturated condition Ef or Vt (V)

Air-gap line OCC

Vrated

SCC

VA

Isc,B

At V = Vrated, Z s , sat =

R +X 2 a

2 s ,sat

=

Vrated (= E f

)

Isc (A)

Isc, A IfA

If (A) IfB

I scB

X s , sat = Z s2, sat − Ra2: Ra is known from the DC test.

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

jXs

Ra Vt=0

+

Ef

Ia Ef

jIaXs

+ Vt=0

Ia

I aR a

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 Ef or Vt (V)

Air-gap line

Isc (A) OCC

Vrated

SCC Isc,rated

I f _ Vrated SCR = I f _ Iscrated =

If_V rated

If_Isc rated

If (A)

1

X s _ sat [in p .u .]

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)

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

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 winding c) 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.

Synchronization Before connecting a generator in parallel with another generator, it must be synchronized. A generator is said to be synchronized 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. a Generator 1

b

Load

c Switch

a/ Generator 2

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. 2. 3. 4.

The rms line voltages of the two generators must be equal. The two generators must have the same phase sequence. The phase angles of two a phases must be equal. 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 a phases. 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

Generat or

Load

Rest of the power system

Xs1 Ef1 Xs2 Ef2

Generato r

G

Xsn Efn

Infinite bus V, f are constant Xs eq = 0

Concept of the infinite bus When a synchronous generator is connected to a power system, the power system is often so large that nothing, the operator of the generator does, will have much of an effect on the power system. An example of this situation is the connection of a single generator to the power grid. Our power grid is so large that no reasonable action on the part of one generator can cause an observable change in overall grid frequency. This idea is idealized in the concept of an infinite bus. An infinite bus is a power system so large that its voltage and frequency do not vary regardless of how much real or reactive power is drawn from or supplied to it.

Steady-state powerangle characteristics

Active and reactive power-angle characteristics

Pm

Pe, Qe Vt

Fig. Synchronous generator connected to an infinite bus.

• 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

Active and reactive power-angle characteristics Pm

Pe, Qe Vt

• 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.

Active and reactive power-angle characteristics Pm

The complex power output of the generator in voltamperes per phase is given by

Pe, Qe Vt

_

S = P + jQ = V t I *a

where: Vt = terminal voltage per phase Ia* = complex conjugate of the armature current per phase Taking the terminal voltage as reference _

V t = Vt + j 0

the excitation( at stator in case of motor) or the generated voltage, _

E f = E f (cos δ + j sin δ )

Active and reactive power-angle characteristics and the armature current, _

_

E f −V t Ia = = jX s _

(E

f

cos δ − Vt ) + jE f sin δ

Pm

jX s

where Xs is the synchronous reactance per phase. _

_

S = P + jQ = V t I

* a = Vt

=

∴ P= Q=

Vt E f sin δ Xs

Vt E f sin δ Xs

&

Vt E f cos δ − Vt2 Xs

(

)

 E f cos δ − Vt − jE f sin δ    jX −   s +j

Vt E f cos δ − Vt2 Xs

Pe, Qe Vt

Active and reactive power-angle characteristics Pm

Pe, Qe Vt

∴ P=

Vt E f sin δ Xs

&

Q=

Vt E f cos δ − Vt2 Xs

• 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-toneutral • 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). Real power or torque Pull-out torque as a generator

−δ

−π

generato r

−π/2 0

+π/2

motor

Pull-out torque as a motor

+π

+δ

Steady-state stability limit Total three-phase power:P =

3Vt E f Xs

sin δ

The above equation shows that the power produced by a synchronous generator depends on the angle δ between the Vt and Ef. The maximum power that the generator can supply occurs when δ=90o. P=

3Vt E f Xs

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

Pull-out torque The maximum torque or pull-out torque per phase that a twopole round-rotor synchronous motor can develop is Tmax =

Pmax Pmax = n ωm 2π s   60 

where ns is the synchronous speed of the motor in rpm P or Q P Q δ

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

BLONDELS TWO REACTION THEORY

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

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 hence considers the results of the cross and directreaction components separately and if saturation is neglected, accounts for their different effects by assigning to each an appropriate value for armature-reaction "reactive" respectively Xaq and Xad . 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 Phenomenon Consider 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), it tends to establish full normal linkage in stator winding. The stator opposes this by a current in the shown direction as to force the flux in the leakage path. The rotor current must increase to maintain its flux constant. It reduces to normal at position (c) where stator current is again reduces to zero. The waveform of stator current and field current shown in the Fig. 5. changes totally if the position of rotor at the instant of short circuit is different. Thus the short circuit current is a function of relative position of stator and rotor. Using the theorem of constant linkages a three phase short circuit can also be studied. After the instant of short circuit the flux linking with the stator will not change. A stationary image of main pole flux is produced in the stator. Thus a d.c. component of current is carried by each phase. The magnitude of d.c. component of current is different for each phase as the instant on the voltage wave at which short circuit occurs is different for each phase. The rotor tries to maintain its own poles

The rotor current is normal each time when rotor poles occupy the position same as that during short circuit and the current in the stator will be zero if the machine is previously unloaded. After one half cycle from this position the stator and rotor poles are again coincident but the poles are opposite. To maintain the flux linkages constant, the current in rotor reaches to its peak value. The stationary field produced by poles on the stator induces a normal frequency emf in the rotor. Thus the rotor current is fluctuating whose resultant a.c. component develops fundamental frequency flux which rotates and again produces in the stator winding double frequency or second harmonic currents. Thus the waveform of transient current consists of fundamental, a.c. and second harmonic components of currents. Thus whenever short circuit occurs in three phase generator then the stator currents are distorted from pure sine wave and are similar to those obtained when an alternating voltage is suddenly 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 its terminals, then the emf induced in the stator winding is used to circulate short circuit current through it. Initially the reactance to be taken into consideration is not the synchronous reactance of the machine. The effect of armature flux (reaction) is to reduce the main field flux. • But the flux linking with stator and rotor can not change instantaneously because of the induction associated with the windings. Thus at the short circuit instant, the armature reaction is ineffective. It will not reduce the main flux. Thus the synchronous reactance will not come into picture at the moment of short circuit. The only limiting factor for short circuit current at this instant is the leakage reactance.

After some time from the instant of short circuit, the armature reaction slowly shows its effect and the alternator then reaches to steady state. Thus the short circuit current reaches to high value for some time and then settles to steady value. It can be seen that during the initial instant of short circuit is dependent on induced emf and leakage reactance which is similar to the case which we have considered previously of voltage source suddenly applied to series R-L circuit. The instant in the cycle at which short occurs also affects the short circuit current. Near zero e.m.f. (or voltage) it has doubling effect. The expressions that we have derived are applicable only during initial conditions of short circuit as the induced emf also reduces after some tome because of increased armature reaction. The short circuit currents in the three phases during short circuit are as shown in the Fig(next slide)

Capability Curves of Synchronous Generators

• The rating of synchronous generators is specified in terms of maximum apparent power in KVA and MVA load at a specified power factor (normally 80, 85 or 90 percent lagging) and voltage for which they are designed to operate under steady state conditions. This load is carried by the alternators continuously without overheating. With the help of automatic voltage regulators the terminal 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 at 0.85 p.f. lagging at rated load. More field current results in overheating of the field system which is undesirable. For this compounding curves of the alternators can be drawn. • If synchronous generator is supplying power at constant frequency to a load whose power factor is constant then curve showing variation of field current versus armature current when constant power factor load is varied is called compounding curve for alternator.

• 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-2 SYNCHRONOUS MOTOR Presented by C.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.

= φR φ= φm sin θ .......................(a) m sin ωt

φ= φm sin (ωt − 120= ) φm sin (θ − 120 ) ...................(b) Y φ= φm sin (ωt − 240= ) φm sin (θ − 240 ) ...................(c) B

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

= φR φ= φ= m sin θ m sin 0 0 3 φY = φm sin (θ − 120 ) = φm sin ( 0 − 120 ) = − φm 2 3 φB = φm sin (θ − 240 ) = φm sin ( 0 − 240 ) = + φm 2

RMF – contd. 

Case (i) - Phasor diagram

RMF – contd.

RMF – contd. 

Case (ii) φ = 60

(look at the waveform)

RMF – contd. 

Simply substitute φ = 60 in equations a, b, c

sin 60 = φR φ= φm= m sin θ

3 φm 2

3 φY = φm sin (θ − 120 ) = φm sin ( 60 − 120 ) = − φm 2 φ= φm sin (θ − 240= ) φm sin ( 60 − 240=) 0 B

RMF – contd.

RMF – contd.

RMF – contd. 

Case (iii) φ = 120

(look at the waveform)

RMF – contd. 

Simply substitute φ = 120 in equations a, b, c

3 = φR φ= φm sin120 = φm m sin θ 2 φ= φm sin (θ − 120= ) φm sin (120 − 120=) 0 Y 3 φR = φm sin (θ − 240 ) = φm sin (120 − 240 ) = − φm 2

RMF – contd.

RMF – contd. 

Case (iv) φ = 180

(look at the waveform)

RMF – contd. 

Simply substitute φ = 180 in equations a, b, c

= φR φ= φm sin180 = 0 m sin θ 3 φ= φm sin (θ − 120= φm ) φm sin (180 − 120=) Y 2 3 φB = φm sin (θ − 240 ) = φm sin (180 − 240 ) = − φm 2

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

Eb = 4.44 K C K d φ fT 

As given, emf is proportional to flux

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

What actually happens ? 



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

What actually happens ? 



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

V=t E f + I a Z s 

It can be expressed as

I= a

Ef Vt − Zs Zs

Excitation Circle Diagram – contd. I= a  

Ef Vt − Zs Zs

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

Excitation Circle Diagram – contd.

Excitation Circle Diagram – contd.  

Same result can be obtained mathematically as follows With Vt as reference

I= a

Ef Vt − Zs Zs

Vt ∠0 E f ∠ − δ = Ia − Z s ∠φ Z s ∠φ

Excitation Circle Diagram – contd. I= a

Ef Vt ∠ −φ − ∠ −δ −φ Zs Zs

Ef  Vt  = cos (δ + φ ) − j sin (δ + φ ) )  I a  ( cos φ − j sin φ ) − ( Zs  Zs 

Re arranging Ef  Vt  = I a  cos φ − cos (δ + φ )  + Zs  Zs 

Ef  Vt  j  − sin φ + sin (δ + φ )  Zs  Zs 

Excitation Circle Diagram – contd. Magnitude 2

V   V  I a2  t cos φ − = cos (δ + φ )  +  − t sin φ + sin (δ + φ )  Zs Zs  Zs   Zs  Ef

2

Ef

2

2

 Vt   E f  Vt E f I =   + cos (δ + φ ) cos φ + sin (δ + φ ) sin φ   −2 Zs Zs  Zs   Zs  2 a

2

2

 V   Ef  Vt E f − 2 I a2 =  t  +  ( cos δ cos φ − sin δ sin φ ) cos φ + ( sin δ cos φ + cos δ sin φ ) sin φ   Z Z Z Z s s  s  s

Excitation Circle Diagram – contd. 2

2

 V   Ef  V Ef I a2 =  t  +   − 2 t ( cos δ cos φ − sin δ sin φ ) cos φ + ( sin δ cos φ + cos δ sin φ ) sin φ  Zs Zs  Zs   Zs  2

2

 V   Ef  V Ef cos δ cos 2 φ − sin δ sin φ cos φ + sin δ cos φ sin φ + cos δ sin 2 φ  I a2 =  t  +   − 2 t Zs Zs  Zs   Zs  2

2

 Vt   E f  Vt E f cos δ cos 2 φ + cos δ sin 2 φ  I =  +   − 2 Zs Zs  Zs   Zs  2 a

2

I a2

2

V  E  V Ef = t  +  f  − 2 t cos δ Z Z Z Z s s  s   s 

Excitation Circle Diagram – contd. 2

2

 Vt   E f  Vt E f cos δ I =  +   − 2 Zs Zs  Zs   Zs  2 a





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

Excitation Circle Diagram – contd. 

Coming back to our diagram (kindly verify the sides)

Excitation Circle Diagram – contd.  

 

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

Excitation Circle Diagram – contd. 

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 Circle Diagram 

Power output per phase is given as

= P Vt I a cos φ − I a2 ra 

P is the mechanical power developed including iron and mechanical losses

Power Circle Diagram  

The equation can be written as, Dividing the whole equation by ra and rearranging it, we get Vt P I − I a cos φ + 0 = ra ra 2 a

Vt P I cos φ + I sin φ − I a cos φ + = 0 ra ra 2 a

2

2 a

2

Power Circle Diagram Vt P I cos φ + I sin φ − I a cos φ + = 0 ra ra 2 a



2

2 a

2

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

Vt P x +y − y+ = 0 ra ra 2



2

This is equation of circle with = centre

 Vt  radius 0,   &=  2ra 

2

 Vt  P −   r ra 2  a 

Power Circle Diagram

Power Circle Diagram



Alternatively, We know,



Vt P I − I a cos φ + = 0 ra ra Adding Vt / 2 ra on either side we get,



2 a

2

Vt P  Vt   Vt  I − I a cos φ + +   =   ra ra  2ra   2ra  2 a

2

Power Circle Diagram Re arranging , 2

2

 Vt   Vt  Vt P I + I cos − = − φ    a r r r ra 2 2 a  a  a 2 a



Slight Modification, yields 2

2

 Vt   Vt  P Vt I + I a cos φ =   −2  − 2ra  2ra   2ra  ra 2 a

Power Circle Diagram 2

2

 Vt   Vt  P Vt 2 Ia +  I a cos φ =   −2  − 2ra  2ra   2ra  ra



The above expression shows that 2

 Vt  P   − ra  2ra 

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

Power Circle Diagram 

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 get 2

 Vt  Pmax = 0   − ra  2ra 

⇒ Pmax

Vt 2 = 4ra

Power Circle Diagram- Inference 

Maximum power input,

 Vt ⇒ Pin ,max = Vt I a cos= φ Vt   2ra 

 Vt 2 =  .1 2ra 

Efficiency is given by 2 V ( Pmax t / 4ra ) = η = = 50% 2 Pin ,max (Vt / 2ra )

Power Circle Diagram- Inference  



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

Power Circle Diagram- Inference

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

Power Developed by Synchronous Motor 

In a motor power developed can be given as

Pm = Eb I a cosψ 

Looking at the phasor diagram again

Power Developed by Synchronous Motor 

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

Power Developed by Synchronous Motor

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 EbV Eb = cos (θ − δ ) − cos θ Pm Zs Zs

Differentiating , dPm EbV = − sin (θ − δ dδ Zs

0 )=

Maximum power developed condition EbV sin (θ − δ − Zs

0 )=

sin (θ − δ ) = 0

⇒θ −δ = 0 ⇒θ = δ

Maximum power developed 

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

EbV Eb2 cos δ − Pm= ,max Zs Zs or Pm= ,max

EbV Eb2 − cos θ Zs Zs

Maximum power developed 

If

Ra ≈ 0 Pm ,max



EbV = Zs

Substituting, cos θ = Ra / Zs Pm= ,max

EbV Eb2  Ra  −   Zs Zs  Zs 

Maximum power developed

Pm= ,max

Solving , Zs E= b 2 Ra

EbV Eb2  Ra  −   Zs Zs  Zs 

V ± V 2 − 4 R ( P  ) a m ,max   

Maximum power developed – condition 

As the equation says, Power developed depends on excitation

EbV Eb2 cos (θ − δ ) − cos θ = Pm Zs Zs 

Differentiating with respect to Eb

2   dPm Eb d EbV = − cos (θ − δ )= cos θ  0  dEb dEb  Z s Zs 

Maximum power developed condition 2   dPm Eb d EbV = cos (θ − δ )= − cos θ  0  dEb dEb  Z s Zs 

Eb

VZ s = 2 Ra

Maximum power developed condition Eb  

VZ s = 2 Ra

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 Pm= ,max

V2 V2 − 2 Ra 4 Ra

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

Operation of AC Generators in Parallel with 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 load 2- having many generators in parallel increase the reliability of power system 3- 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

Operation of AC Generators in Parallel with Large Power Systems 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)

Operation of AC Generators in Parallel with Large Power Systems • When a syn. Gen. connected to a power system: 1-The real power versus frequency characteristic of such a system 2-And the reactive power-voltage characteristic

Operation of AC Generators in Parallel with Large Power Systems • Behavior of a generator connected to a large system A generator connected in parallel with a large system as shown • Frequency & voltage of all machines must be the same, their real powerfrequency (& reactive power-voltage) characteristics plotted back to back 

Operation of AC Generators in Parallel with Large Power Systems • 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

Operation of AC Generators in Parallel with Large Power Systems • In order that a generator comes on line and supply 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 motors 2. Using damper winding 3. As a slip ring induction motor 4. 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 Motor Refer 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. These machines are made self starting by providing a special winding in the rotor poles, known as damper winding or squirrel cage windings. The damper winding consists of short circuited copper bars embedded in the face of the rotor poles When an ac supply is provided to stator of a 3-phase synchronous motor, stator winding produces rotating magnetic field. Due to the damper winding present in the rotor winding of the synchronous motor, machine starts as induction motor (Induction machine works on the principle of induction. Damper windings in synchronous motor will carryout the same task of induction motor rotor windings. Therefore due to damper windings synchronous motor starts as induction motor and continue to accelerate). The exciter for synchronous motor moves along with rotor. When the motor attains about 95% of the synchronous speed, the rotor windings is connected to exciter terminals and the rotor is magnetically locked by the rotating magnetic field of stator 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 starting torque • By providing damper windings in the rotor of synchronous motor "Hunting of machine“ can be suppressed. When there is change in load, excitation or change in other conditions of the systems rotor of the synchronous motor will oscillate to and fro about an equilibrium position. At times these oscillations becomes more violent and resulting in loss of synchronism of the motor and 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.

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

• 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-3 Three phase Induction Motor

Presented By C.GOKUL AP/EEE

UNIT 3 Syllabus

Construction of Induction Motor

Types of Rotor

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

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

SLIP(s)

Compare Induction motor & Transformer

Equivalent circuit

Losses & Efficiency

Losses - Summary

Efficiency (η) =

Poutput Pinput 356

Motor Torque Tm =

9.55 Pm n 9.55 (1 – s) Pr

= ns (1 – s) = 9.55 Pr / ns Tm = 9.55 Pr / ns

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

357

I2R losses in the rotor Pjr = s Pr Pjr = rotor I2R losses [W] s = slip Pr = power transmitted to the rotor [W]

Mechanical Power Pm = Pr - Pjr = Pr - s Pr = (1 – s) Pr 358

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 or Open Circuit Test This test gives 1. Core loss 2. F & W loss R 3. No load current I0 4. No load power factor STATOR 5. Ic, Rc, Iμ, Xm 6. Mechanical faults, noise

W0

I0

A N V0 V B

Y ROTOR

Rated per voltage V0, with rated freq is given to stator. Motor is run at NO LOAD P0, I0 and V0 are recorded P0 = I02r1+Pc+Pfw

P0 Cosθ0 = V0 I0

1. Ic=I0cosθ0

No load power factor is small, 0.05 to 0.15 provided x1 is known

2. Iμ=I0sinθ0

I0

Ic V0

Rc

IΦ jXm

open circuit

E0 E0 , E 0 =V 0 − I 0 (r 1 + jx 1) 3. R c = 4. X m = Ic Iµ On No load, Motor runs near to syn speed So, s ≈ zero 1/s=α or open circuit jx2 r1 I jx1 I 2 0

r2/s

The F & W loss Pfw, can be obtained from this test. Vary input voltage and note input power Input Power

Pfw Input Voltage Thus Pc=P0 - I02r1 - Pfw Presented by C.GOKUL,AP/EEE Velalar College of Engg & Tech , Erode

BLOCKED ROTOR TEST

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

A

R

N

IM

V Vsc B

3-ph Variac

Y Rotor is blocked or held stationary by belt pulley or by hand

Low voltage is applied upto rated stator current Voltage Vsc, Current Isc and Power Psc are measured.

Since slip is 1, secondary is short circuited r1

jx1 Ic

Vsc

Isc I0

Rc

This test gives copper loss Mechanical loss =0 Core loss negligible Rc and Xm >> r2+jx2 Hence omitted

r2

IΦ jXm

jx2

1 −s  r2   s  

Therefore, Zsc = Vsc / Isc =Rsc+jXsc P sc cosθsc = =0.8 to 0.9 VscI sc

Rsc= Psc/Isc2 = r1+r2

r2= Rsc – r1

X sc= Z sc −R sc 2

2

=x 1+x 2

For wound rotor motor, x1 = x2 = Xsc /2 For squirrel cage motor, Class of motor

x1

x2

1.

Class A (normal Tst and Ist)

0.5

0.5

2.

Class B (normal Tst and low Ist)

0.4

0.6

3.

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

Circle Diagram of Ind Motor

Graphical representation

The equivalent ckt., operating ch. can be obtained by computer quickly and accurately But the advantage of circle diagram is that extremities or Limits of stator current, Power, torque and slip can be known from circle diagram The circle diagram is constructed with the help of 1. No load test (I0 & θ0) 2. Blocked rotor test (Isc & θsc)

y Isc

V1

Output line

θsc O

θ0 I 0

x 4. Join I0 and Isc

1. Draw x and y axes(V1 on y axis) Line I0Isc is 2. Draw I0 and Isc(=V1/Zsc) output line 3. Draw parallel line to x axis from I0. This line indicates constant loss vertically

y Isc

V1

Output line T

θsc θ0 I 0

C

L1

O L2 5. Draw perpendicular bisector to output line

x

6. Draw circle with C as a centre 7. Draw perpendicular from Isc on x axis.. I scT r2' Rotor Cu Loss = = 8. Divide IscL1 in such a way that. T L1 r1 Stator Cu Loss

R

y

rated output power Isc

V1

Output line T Torque line

θsc O

θ0 I 0

C

L1 L2

x

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 Total cm for rated output power=IscR

R

y P’ V1 P θ1 θsc

Isc Output line T Torque line

O’ T’ L1 θ0 I ’ L 1 0 C x O L2’ L2 11. From R, draw line parallel to output line crossing at P & P’. P is operating point 12. Join O and P. Cosθ1 is operating pf. 13. From P draw perpendicular on x axis Lebel O’, T’ , L1’ and L2’

R

y P’

Isc

V1 P

Output line

θ1 θsc O

θ0 I 0

O’ T’ L1’ L2’

T Torque line C

L1 L2

14. Determine the following 1. Constant Losses and copper losses L1L2=L1’L2’=constant losses =Core loss + F & W loss α no load current I0

x

R

y P’

Isc

V1 P

Output line

θ1 θsc O

θ0 I 0

O’ T’ L1’ L2’

At standstill, input power = IscL2

T Torque line C

L1 L2

L1L2=Constant Loss

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

x

R

y P’

Isc

V1 P

Output line

θ1 θsc O

θ0 I 0

O’ T’ L1’ L2’

T Torque line C

L1 L2

x

At operating point P, input power = PL2’, L1’L2’=Constant Loss Constant loss = Stator core loss + F & W loss Rotor core loss ≈ 0 (sf) Thus L1L2=L1’L2’= Constant loss

R

y P’

Isc

V1 P

Output line

θ1 θsc O

θ0 I 0

O’ T’ L1’ L2’

At standstill, Stator Cu loss=TL1 At P, stator Cu loss =T’L1’ and

T Torque line C

L1 L2

rotor Cu loss = IscT

rotor Cu loss = O’T’

x

R

y Pmax

V1 P θ1

P’

Isc Output line T Torque line

O” O’ T’ T” L1 θ0 I L ” ’ L 1 1 0 C O L2’ L2” LP2max 2. Output Power and Torque Output Power = O’P The gap betn output line and circle is OUTPUT Power. 0 Speed At I0, o/p=0, at Isc, o/p=0 1 Slip Max output power=PmaxO” θsc

x

Ns 0

R

y Pmax

V1 P θ1

P’ Tmax

O”’

Isc Output line T Torque line

O” O’ T’ T” T”’ L1 L1”’ θ0 I ” L ’ L 1 1 0 C x L ”’ O L2’ L2” 2 L2 Tmax 2. Output Power and Torque Output Torque = T’P The gap betn torque line and circle is OUTPUT torque. Ns 0 Speed At I0, torque=0, but at Isc, torque=T Isc=Starting torque 1 Slip 0 Max output torque=TmaxT”’ θsc

R

y Pmax

V1 P θ1

P’ Tmax

O”’

O” O’ T’ T” T”’ L1”’ θ0 I ” L ’ L 1 1 0 C L ”’ O L2’ L2” 2 2. Output Power and Torque θsc

Isc Output line T Torque line L1 L2

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

x

R

y Pmax

V1

θsc

O’ T’ θ0 I L1’ 0 O L2’ 3. Slip, Power factor and

Isc

Tmax

P θ1

P’

O”

Output line

O”’

T Torque line

T”’ T” L1” L1”’ C L ”’ L2” 2 Efficiency

L1 L2

Air gap power Pg = Input power – Stator Cu loss- core loss =PL2’-T’L1’-L1’L2’ = PT’ s = rotor Cu loss/Pg =O’T’/PT’

smp

O"T " = Pmax T "

x

O" ' T " ' smt = Tmax T " '

R

y Pmax

V1

Tmax

P θ1 θsc

O’ T’ θ0 I L1’ 0 O L2’ 3. Slip, Power factor and

P’

O”

O”’

T”’ T” L1” L1”’ C L ”’ L2” 2 Efficiency

Power factor cosθ1 = PL2’/OP Efficiency= PO’/PL2’

Isc Output line T Torque line L1 L2

x

R

y Pmax

V1

Tmax

P θ1 θsc O

θ0 I 0

4. Braking Torque

P’

O’ T’ L1’ L2’

O”

Isc Output line braking torque T Torque line

T” C

s=1 Te L 1 L2

x

s=α Speed s=0 0 Ns 1 The gap betn circle and T & s=α is braking Slip 0 α torque

R

y Pmax

V1

Tmax

P θ1 θsc

O’ T’ θ0 I L1’ 0 O L2’ 5. Induction Generator s=0

P’

O”

Isc Output line braking torque T Torque line s=1

T” C

L1 L2

x s=α

R

y Pmax

V1

P’ Tmax

P

Isc Output line braking torque

θ1

O” O’ T” T’ θ0 I L1’ 0 C O L2’ 5. Induction Generator s=0 θG s= -ve G (Generator) θsc

PGmax

T Torque line s=1

L1 L2

x s=α

OG=Gen Current O’G=Mech I/p L2’G=Active power

OL2’=reactive power

R

y Pmax

V1

P’ Tmax

P θ1

Isc Output line braking torque T Torque line

O” O’ s=1 T” T’ L1 θ0 I ’ L 1 Te 0 C x O L2’ L2 5. Induction Generator s=α Speed Speed s=0 0 Ns 2Ns θG OG=Gen Current 1 Slip 0 Slip -1 α O’G=Mech I/p s= -ve G (Generator) L2’G=Active power θsc

PGmax

OL2’=reactive power

CIRCLE DIAGRAM OF AN INDUCTION MOTOR- Summary T

H

Fig. 3.3

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

SEPARATION OF NO LOAD LOSSES The separation of core loss and mechanical loss (windage and friction) can be obtained by no load test conducted from variable voltage, rated frequency supply. Step by step reduce the 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 loss remains 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 graph Magnetic loss + mechanical loss = output power Therefore., magnetic loss = output power – mechanical loss

Formulae for calculating the equivalent circuit parameters: Z0 = Voc /(Ioc / √3) R0 = Woc / (Ioc) 2 X0 = √[( Z0)2 - (R0)2 ϕ0 = cos-1 [Woc / (√3 * Voc * Ioc )] RBR = Wsc / (Isc)2 ZBR = Vsc / (Isc/ √3) XBR = √[( ZBR)2 - (RBR)2] RiWF – Resistance accounting for rotational losses R1 = 1.2 * stator winding resistance (dc) Pr = Woc – Ioc2 * R1 (since Pr = P0 – 3 * (Ioc / √3)2 * R1) RiWF = Voc2 / Pr Xm – Magnetizing reactance IiWF = Voc / Riwf Im = (Ioc2 - 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.

Double Cage construction Fig. 3.5

Slip Torque-Slip Characteristics

Equivalent Circuit:

‘

‘

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

Rotor

Induction Generators

INDUCTION GENERATOR

Principle of operation Induction generators and motors produce electrical power when their rotor is rotated faster than the synchronous speed. For a fourpole 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.

a. Sub-synchronous (motor)

b. Super-synchronous (generator)

Fig. 3.4 current Locus for Induction Machine

Fig.3.5 Phasor Diagram

Fig. 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 current AB-stator current to overcome rotor mmf OB-total stator current

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. Fig.3.4b the point P in the lower half of the circle shows operating point as an induction generator. PT-stator electrical output ST-Core, friction and windage losses RS-Stator copper loss QR-Rotor copper loss PQ-Mechanical input PR-Rotor input Slip Efficiency

=

rotor copper loss QR = rotor input PR

output PT = = input PQ

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 MOTOR It 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. Stator

3Φ Supply

Fig. 3.3

Rotor connections for dc excitation:

Fig 3.4 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.

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:

120 f

120 * 60 ns = = = 1800 rpm p 4 n = (1 − s )ns = (1 − 0.05) *1800 = 1710 rpm (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 V No-load current: 14 A No-load power: 1470 W Resistance measured between two terminals: 0.5 Ω The locked-rotor test, conducted at reduced volt­age, gave the following results: Locked-rotor voltage (line-to-line): 163 V Locked-rotor power: 7200 W Locked-rotor current: 60 A Determine the equivalent circuit of the motor. Solution: Assuming the stator windings are connected in way, the resistance per phase is:

R1 = 0.5 / 2 = 0.25 Ω From the no-load test:

VLL 440 V1 = = = 254 V / Phase 3 3

Z NL

V1 254 = = = 18.143 Ω I1 14

R NL =

PNL 3I12

1470

=

3 *14

2

= 2.5 Ω

2 2 X NL = Z NL − RNL = 18.1432 − 2.52 = 17.97

X 1 + X m = X NL = 17.97 Ω From the blocked-rotor test

RBL =

PBL 3I12

=

7200 3 * 60

2

= 0.6667 Ω

BL

The blocked-rotor reactance is:

X BL =

(Z

2 BL

)

2 − RBL = 1.5685 2 − 0.6667 2 = 1.42 Ω

X BL ≅ X 1 + X 2′ = 1.42 Ω

∴ X 1 = X 2′ = 0.71 Ω

X m = X NL − X 1 = 17.97 − 0.71 = 17.26 Ω R = RBL − R1 = 0.6667 − 0.25 = 0.4167 Ω  X 2′ + X m ∴ R2′ =   Xm

2

2  0 . 71 17 . 26 +   R =   * 0.4167 = 0.4517 Ω   17.26  

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 V Line current = 4.5 A, Input power = 1600 W (2) Blocked-rotor test: Frequency = 15 Hz, Line voltage = 270 V Line 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.

2200 V1 = = 1270.2 V / Phase 3

RNL

Z NL

V1 1270.2 = = = 282.27 Ω 4.5 I1

PNL 1600 = 2 = = 26.34 Ω 2 3I1 3 * 4.5

(a) No-Load equivalent Circuit

(b) Locked rotor equivalent circuit

2 2 X NL = Z NL − RNL = 282.27 2 − 26.34 2 = 281Ω

X 1 + X m = X NL = 281 Ω PBL 9000 RBL = 2 = = 4.8 Ω 2 3I1 3 * 25

R2′ = RBL − R1 = 4.8 − 2.8 = 2Ω

281.0 = Ω.

impedance at 15 Hz is:

Z BL

270 V1 = = = 6.24 Ω I1 3 * 25

The blocked-rotor reactance at 15 Hz is Its value at 60 Hz is

X BL = 3.98 *

X BL =

(6.24

2

60 = 15.92 Ω 15

X BL ≅ X 1 + X 2′ 15.92 ∴ X 1 = X 2′ = = 7.96 Ω 2

X m = 281 − 7.96 = 273.04 Ω R = RBL − R1 = 4.8 − 2.8 = 2 Ω

7.96 + 273.04   R2′ =   2 = 2.12 Ω  273.04  2

)

− 4.82 = 3.98 Ω

at 60 Hz

)c (

273.04 Vth ≅ V1 = 0.97 V1 7.96 + 273.04

Rth ≅ 0.97 R1 = 0.97 * 2.8 = 2.63 Ω 2

2

X th ≅ X 1 = 7.96 Ω

Example 4.4 A three-phase, 460 V, 1740 rpm, 60 Hz, four-pole wound-rotor induction motor has the following parameters per phase:

R1 = 0.25 Ω, R2′ = 0.2 Ω, X 1 = X 2′ = 0.5 Ω, X m = 30 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

28022.3 ηmotor = *100 = 87.5% 32022.4 ηint ernal = (1 − s ) *100 = (1 − 0.0333) *100 = 96.7% (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:

(a ) Z NL = RNL =

460 / 3 = 6.64 Ω 40

PNL 2 3 * I1

4200

=

3 * 40

2

= 0.875Ω

X NL = 6.64 2 − 0.8752 = 6.58Ω

X 1 + X m = 6.58Ω From blocked rotor test:

RBL = Z BL =

8000 3 *140 2

= 0.136Ω

100 / 3 = 0.412Ω 140

0.152 R1 = = 0.076Ω 2

X BL = 0.412 2 − 0.136 2 = 0.389Ω

X 1 + X 2′ = 0.389

0.389 = 0.1945Ω X 1 = X 2′ = 2 X m = 6.58 − 0.1945 = 6.3855

R = RBL − R1 = 0.136 − 0.076 = 0.06Ω 0.1945 + 6.3855   R2′ =   * 0.06 = 0.0637 6.3855   2

0.076 Ω

j0.195 Ω

j6.386 Ω

j0.195 Ω

0.0637 s

(b )

120 f 120 * 60 ns = = = 900rpm P 8

ns − n 900 − 873 = = 0.03 s= 900 ns R2′ 0.0637 = = 2.123 s 0.03 Input impedance

Z1 = 0.076 + j 0.195 +

( j 6.386)(2.123 + j 0.195) = 2.121∠27.16o Ω 2.123 + j (6.386 + 0.195)

V1 460 / 3 o = = 125.22∠ − 27.16 I1 = Z1 2.12∠27.16 Input power:

(

)

460 Pin = 3 * *125.22 cos 27.16o = 88.767 kW 3

Stator CU losses:

Pst = 3 *125.22 2 * 0.076 = 3.575 kW Air gap power

Pag = 88.767 − 3.575 = 85.192 kW

Rotor CU losses

P2 = sPag = 0.03 * 85.192 = 2.556 kW

Mechanical power developed:

Pmech = (1 − s ) Pag = (1 − 0.03) * 85.192 = 82.636 kW

Pout = Pmech − Prot From no load test:

Prot = PNL − 3I12 * R1 = 4200 − 3 * 40 2 * 0.076 = 3835.2 W

Pout = 82.636 *103 − 3835.2 = 78.8 kW

Pout 78.8 η= *100 = *100 = 88.77 % 88.767 Pin

Example

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

wound rotor induction motor has the following parameters per phase ( R1 =0.2Ω, R2′ =0.18 Ω, X 1 = X 2′ =0.2Ω, X m =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?

460 V1 = = 265.6 V / phase 3

j 40 * (0.18 + 0.2 ) o = 0.55∠46.59 Ω Z1 = 0.2 + j 0.2 + 0.18 + j 40.2 V1 265.6 o = 482.91 ∠ − 46.3 Ω I st = = o I1 0.55∠46.59 1500 − 1450 s= = 0.0333 1500 R2′ 0.18 = = 5.4 s 0.0333

j 40 * (5.4 + j 0.2 ) = 4.959 ∠10.83o Ω Z1 = 0.2 + j 0.2 + 5.4 + j 45.4

I1 FL

265.6 o = = 53.56∠ − 10.83 A o 4.959∠10.83

Then the power factor is: cos 10.83 = 0.9822 lag. o

ω sys

1500 = * 2π = 157.08 rad / sec . 60

265.6 * ( j 40 ) = 264.275 ∠0.285o V Vth = (0.2 + j 40.2) Then,

j 40 * (0.2 + j 0.2 ) = 0.281432 ∠45.285o = 0.198 + j 0.2 Ω Z th = 0.2 + j 40.2

∴T =

3 * (264.275) * 5.4 2

157.08 * (0.198 + 5.4 ) + (0.2 + 0.2 ) 2

2

= 228.68 Nm

Then, Pag = T * ω sys = 228.68 *157.08 = 35921.1W Then, P2 = sPag = 0.0333 * 35921.1 = 1197 W And, Pm = (1 − s )Pag = 34723.7W Then, Pout = Pm − Prot = 34723.7 − 1500 = 33223.7W

Pin = 3 * 265.6 * 53.56 * 0.9822 = 41917 W Then, η =

∴ Tm =

Pout 33223.7 = = 79.26 % 41914 Pin 3 * (264.275)2

[

(

2 *188.5 0.198 + 0.1982 + (0.2 + 0.2 )

sTmax =

[0.198

0.18 2

)]

2 1/ 2

+ (0.2 + 0.2 )

]

2 1/ 2

= 862.56 Nm

= 0.4033

(d) sTmax = 1 =

[0.198

′ R2′ + Rext 2

+ (0.2 + 0.2 )

]

2 1/ 2

′ = 0.446323 Then, R2′ + Rext ′ = 0.446323 − 0.18 = 0.26632 Ω Then, Rext

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

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

2 2

2 2

I R2 I R2 α T= sω syn s

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:

Tst Tst Tmax 1.5 = = 0.75 = 2 then, = 1.5 and TFL TFL Tmax 2 2 sTmax Tst = = 0.75 2 Tmax 1 + sTmax Then,

2 0.75 sTmax

− 2 sTmax + 0.75 = 0

Then sTmax = 2.21525 (unacceptable) Or sTmax = 0.451416

2 sT2max + s FL

Tmax = =2 TFL 2sTmax * s FL But sTmax = 0.451416 2 Tmax 0.4514162 + s FL = =2 Then TFL 2 * 0.451416 * s FL 2 s FL − 4 * 0.451416 s FL + 0.451416 2 = 0 2 s FL

− 1.80566 s FL + 0.203777 = 0

s FL = 1.6847 (unacceptable) or s FL = 0.120957

120 * 50 ns = = 1500 rpm 4 then (a) nFL = (1− s FL ) * ns

nFL = (1 − 0.120957 ) *1500 = 1319 rpm

(

)

(b) nTmax = 1 − sTmax * ns = (1 − 0.451416) *1500 = 823 rpm

Example 4.10 A 3φ, 280 V, 60 Hz, 20 hp, four-pole induction motor has the following equivalent circuit parameters.

R1 = 0.12 Ω, R2′ = 0.1 Ω, X 1 = X 2′ = 0.25 Ω, and X m = 10 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. Solution:

120 * 60 1800 ns = * 2π = 188.5 rad / sec = 1800 rpm , ω s = 4 60

0.12 Ω

j0.25 Ω

j0.25 Ω

j10 Ω

0.1 =2 0.05

Z1 = 0.12 + j 0.25 + Re + X e j10 * (2 + j 0.25) Z1 = 0.12 + j 0.25 + = 2.1314∠23.55o Ω 2 + j10.25 V1 =

208

= 120.1 V

3

120.1 o I1 = = 2 . 1314 ∠ − 23 . 55 A o 2.1314∠23.55 (c) P1 = 3 * 56.3479 * 0.12 = 1143.031W 2

(

(d) Ps = 3 *120.1 * 56.3479 * cos − 23.55

o

) = 18610.9794 W

Pag = Ps − P1 = 17467.9485 W (e) P2 = sPag = 0.05 *17467.9785 = 873.3974 W (f) Pm = (1 − s ) Pag = 16594.5511W

Pag

17467.9485 (g) T = = = 92.6682 N .m 188.5 188.5 Pshaft

16194.5511 Tshaft = = = 85.9127 Nm 188.5 188.5 Pshaft (h) η = *100 = 87.02% Ps

Example 4.11 A 30, 100 WA, 460 V, 60 Hz, eight-pole induction machine has the following equivalent circuit parameters:

R1 = 0.07 Ω, R2′ = 0.05 Ω, X 1 = X 2′ = 0.2 Ω, and X m = 6.5 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:

Vth =

Xm 6.5 * V1 = * 265.6 = 257.7 V X1 + X m 0.2 + 6.5

Rth + jX th =

( j 6.5) * ( j 0.2 + 0.07 ) = 0.06589 + 0.07 + j 0.2 + j 6.5

0.06589Ω j0.1947 Ω

j0.2 Ω

257.7V

(b) Tst =

Tmax =

0.05 s 3 * 257.7 2 * 0.05

[

94.25 (0.06589 + 0.05) + (0.1947 + 0.2 ) 2

2

]

= 624.7 Nm

3 * 257.7 2

[

2 * 94.25 0.06589 + 0.06589 2 + (0.1947 + 0.2 )2

= 2267.8 Nm sTmax =

j 0.1947 Ω

0.05 0.06589 2 + (0.1947 + 0.2 )2

= 0.1249

]

Speed

in

(

)

rpm

for

which

max

torque

= 1 − sTmax * ns = (1 − 0.1249 ) * 900 = 787.5 rpm (c) sTmax = or R2′

start

=

R2′

R12

+ ( X 1 + X 2′ )

s start = 1 sTmax

2

* R2′ =

1 0.1249

α R2′ * 0.05 = 0.4 Ω

Then Rext = (0.4 − 0.05) / 1.2 2 = 0.243 Ω

occurs

UNIT-4 Starting & Speed control of 3ph Induction Motor Presented by C.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. this mean that 3-pahse induction motor is a self-starting motor and begins to rotate on its own when connected to a 3-phase supply. At the instant of starting 3-phase induction motor behaves like a transformer with a short-circuited secondary. Consequently, a 3-pahse induction motor takes high starting current if started at full voltage. In order to limit this high starting current to reasonable limits starting 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

V1 IL

xV1

xV1 I st = xI sc

Rotor Stator

Fig: Auto-transformer starting The fraction of xV1 is applied to the stator wdg at starting. As speed increases, gradually voltage is increased Finally full voltage is applied to the motor. Advantages 1. Voltage is changed by transformer action and not by dropping voltage as that of reactor 2. So power loss and input current are less.

V1 IL

xV1

xV1

Rotor Stator

I st = xI sc

Fig: Auto-transformer starting The stator starting current is I st = xV1 / z1 = xI sc For auto-transformer, input VA= output VA ILV1=Ist (xV1) Therefore, line current at IL=xIst input is x2 times the DOL current. IL=x2Isc Thus,

Test Tefl

 I1st =   I1fl

2

  sfl  

 2 I sc = x   Ifl

2

  sfl  

V1 IL

xV1

xV1

Rotor Stator

I st = xI sc

Fig: Auto-transformer starting Line current at input due to auto-transformer starting =x Line current at input due to stator reactor starting V1 IL

xV1

xV1 I st = xI sc

Rotor Stator

V1 IL

xV1

xV1 I st = xI sc

Rotor Stator

Fig: Auto-transformer starting Line current at input due to auto-transformer starting =x Line current at input due to stator reactor starting Starting torque with auto transformer starting 2 =x Starting torque with DOL starting Starting torque with auto transformer starting =1 Starting torque with stator reactor starting

STAR DELTA STARTER

Star-Delta starting For star, 3 terminals of stator wdg are required. For delta, 6 terminals are required. Now make delta Connection. R Y B

Stator 2- Run - Delta TPDT 1- Start - Star

Rotor

Fig.: Star-Delta starting

At starting TPDT to 1, wdg in star Reduced voltage is applied to wdg = VL/√3 Motor rotates. The starting current is Now TPDT to 2- Delta I st.y =VL / 3z 1 Line voltage applied R Y B Starting = I L.y to wdg. Motor runs at rated speed Line current Stator 2- Run - Delta TPDT 1- Start - Star

Rotor

Fig.: Star-Delta starting

At starting TPDT to 1, wdg in star Reduced voltage is applied to wdg = VL/√3 Motor rotates. The starting current is Now TPDT to 2- Delta I st.y =VL / 3z 1 Line voltage applied R Y B Starting = I L.y to wdg Motor runs at rated speed Line current Stator 2- Run - Delta TPDT 1- Start - Star

Rotor

Fig.: Star-Delta starting

At starting TPDT to 1, wdg in star Reduced voltage is applied to wdg = VL/√3 Motor rotates. The starting current is Now TPDT to 2- Delta I st.y =VL / 3z 1 Line voltage applied Starting = I L.y to wdg Motor runs at rated speed Line current At starting, if, wdg in delta The starting current is I st.d =VL / z 1 = I sc.d I L.d = 3 I st.d 1 ∴ I st.y = I st.d 3

Ist.y Starting line current with Y-Δ starter = Starting line current with stator in Δ √3 Ist.d

=

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

1 3

(V1/√3)2 Starting torque with Y-Δ starting = = 2 Starting torque with stator in Δ V1

1 3

This shows that Tst.y in star is one third of starting torque in delta. In case of auto-transformer, if turn ratio x = 1/√3 Then starting line current and is starting torque are reduced to one third of their values with delta. This shows that Star delta starting is equivalent to auto transformer if auto transformer turn ratio x=1/√3=0.58 or 58% tapping This method is cheap, effective and used extensively Used for tool drives, pumps, motor-generator set. Used up to rating of 3.3kV, After this voltage, m/c becomes expensive for delta winding

Example Determine the % tapping of the auto-transformer so that the supply current during starting of IM does not exceed 1.5 times full load current. The short circuit current on normal voltage is 4.5 times the full load current and the full load slip is 3%. Calculate the ratio of starting torque full load torque. Solution V1

IL=1.5IFL Isc=4.5IFL IL/Isc=0.333

IL

xV1

xV1

In auto-transformer

I st = xI sc

IL/Isc=x2

Hence % tapping is 57%

Rotor Stator x=0.577

Test

Now

Tefl

 I1st =   I1fl

2

  2 I sc  sfl = x  I   fl 

2

  sfl  

= 0.333 (4.5 ) 0.03 2

= 0.202

V1 IL

xV1

xV1 I st = xI sc

Rotor Stator

Example The short circuit line current of a 6hp IM is 3.5 times its full load current, the stator of which is arranged for star delta starting. The supply voltage is 400V, full load effn is 82% and full load power factor is 0.85% (lag). Calculate the line current at the instant of starting. Neglect magnetizing current. Solution P=√3 VLILcosθ 6hp IM, 1  6 ×746  Isc=3.5IFL IFL= I L =    0.82  3 × 400 ×0.85 Star-delta starting =9.26A (line current for Isc (line) =3.5 IFL delta) Voltage =400V =5.34A (phase current for delta) η =82%, pf=0.85 (lag) FL

Isc=3.5IFL=3.5x5.34 =18.73A At the instant of starting, motor wdg is in star For 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

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

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 and design. Such large current of short duration don’t harm the rugged squirrel cage motor, but the high currents may cause objectionable voltage drop in power supply feeding the induction motor These large voltage drop causes undesirable dip in the supply line voltage, consequently affecting the other 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:1 2 r2 Te = .I 2 . ωs s Therefore:2 2 Ts I s r2 1  I s  .s f = 2 =   ------Eqn(1) T f I f r2 s f  I f  Now

V1 I st = = I sc Z sc



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



Therefore Eqn(1) can be written as:2



Ts  I sc  .s f =   Tf  I f  ----Eqn(2)

Zsc

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 a fraction ‘x’ of supply voltage appears across it.  This reduces the per phase starting currents Is drawn by the motor from the supply mains.  As the motor speeds up, the reactor is cut out in steps and finally short-circuited when the motor speed is near to synchronous speed. 



Since the per phase voltage is reduced to ‘xV1’ the per phase starting current is:xV1 Is = = xI sc Z sc



Now we know:-



Therefore we have:-

1 I 22 r2 . T= ωs s 2

 

Therefore:-

  Ts 2  I sc  sf =x   -----Eqn(1) Tf  If  2

starting torque with reactor starting  xV1   = x 2 =  starting torque with direct switching  V1 

Rotor resistance Starter

ROTOR RESISTANCE STARTER(only for slip ring induction motor)

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

• 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 ' r

2

2

3R Vs Te = ' 2 sωs   Rr  2  Rs +  + ( X ls + X lr )  s   

3

2 4 ωs = ω = πf P P

1

Varying voltage (amplitude)

Varying line frequency

Pole Changing 501

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 ∝ sE22 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 because1) 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 decreases Te ∝ (1/ωs)

 For control below base 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 : E1 Vs φag = ≈ f f

Assuming small voltage drop across Rs and Lls

 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

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

E1 Vs φag = ≠ f f

Tmax ∝

Vs

2

ωs

 For above base speed operation (f > frated):  stator voltage maintained at rated value  Same as Variable Frequency control (refer to slide 13)

508

Constant Volts/Hz (V/f) Control Vs

Vs vs. f relation in Constant Volts/Hz drives Boost - to

compensate for voltage drop at Rs and Lls

Vrated

Linear offset curve – • for high-starting torque loads • employed for most applications

Linear offset

Boost

Non-linear offset curve – • for low-starting torque loads

Non-linear offset – varies with Is frated

f 509

Constant Volts/Hz (V/f) Control • For operation at frequency K times rated frequency:

– fs = Kfs,rated ⇒ ωs = Kωs,rated (1) (Note: in (1) , speed is given as mechanical speed)

 KVs ,rated , when f s < f s ,rated – Stator voltage:Vs =  (2)  Vs ,rated , when f s > f s ,rated

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

Vs,rated

ω s,rated

(3) 510

Constant Volts/Hz (V/f) Control  For operation at frequency K times rated frequency:

Hence, the torque produced by the motor: Te =

' r

2

Vs 3R ' 2 sω s   Rr  2 2  Rs +  + K ( X ls + X lr )  s   

(4)

where ωs and Vs are calculated from (1) and (2) respectively. 511

Constant Volts/Hz (V/f) Control  For operation at frequency K times rated frequency:

The slip for maximum torque is: smax = ±

Rr'

(5)

Rs + K 2 ( X ls + X lr ) 2

2

 The maximum torque is then given by: Tmax =

Vs

3

2

2ω s  R ± R + K 2 ( X + X )  s ls lr  s  2

2

(6)

where ωs and Vs are calculated from (1) and (2) respectively. 512

Constant Volts/Hz (V/f) Control Rated (Base) frequency

Constant Torque Area

(below base speed)

Field Weakening Mode (f > frated)

• Reduced flux (since Vs is constant) • Torque reduces

⇒Constant Power Area (above base speed)

Note: Operation restricted between synchronous speed and Tmax for motoring and braking regions, i.e. in the linear region of the torque-speed curve. 513

Constant Volts/Hz (V/f) Control Constant Torque Area

Constant Power Area 514

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 Hz i) synchronous speed when 4 pole winding is connected, Ns = 120*50/4 = 1500 RPM ii) synchronous speed when 6 pole winding is connected, Ns = 120*50/6 = 1000 RPM

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

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 B P1 = number of poles stator of motor A P2 = number of stator poles of motor B N = speed of the set and same for both motors f = 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 emf therefore, 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 obtained 1. when only motor A works, corresponding speed = Ns1 = 120f / P1 2. when only motor B works, corresponding speed = Ns2 = 120f / P2 3. 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 RYB

Voltage regulating device

f MIM

ACM

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

2) Scherbius System RYB

RYB f MIM

Voltage regulating device

ACM

AIM

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

Constant torque drive

Braking of 3ph Induction Motors •Plugging •Dynamic Braking •Regenerative Braking

Braking Methods • • •

Regenerative Braking Plugging or reverse voltage braking Dynamic ( or rheostatic ) braking : a) b) c) d)

ac dynamic braking Self-excited braking using capacitor dc dynamic braking 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

B

TL A

Te

8 poles 4 poles

D

0 Speed 1 Slip C

TL

D A Ns 0 B

-Te

+Te

Two quadrant operation

Regenerative braking • Power input to induction motor: Pin=3VIscosφs Motoring operation φs<90º Braking φs>90º ωm> ωms

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.

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

•

From the figure beside we can see that the torque is not zero at zero speed. That’s why when the motor is needed to be stopped, it should be disconnected from the supply at near zero speed. • The motor is connected to rotate in the reverse direction and the torque is not zero at zero or any other speed, and as a result the motor first decelerates to zero and then smoothly accelerates in the opposite direction.

3. DC Dynamic Braking or Rheostatic or AC Dynamic Braking The disadvantages of plugging are removed in dynamic braking. Dynamic braking requires less power. Under normal operating condition Stator - Rotating Magnetic Field - Ns

Faster sNs

Rotor - Te -

Slower

Rotates - Nr

If DC supply is given to stator Stator - Stationary Magnetic Field -Ns =0

Slower

Rotor - Teb Rotates - Nr Faster Ns(1-s)= NsS This Teb is dynamic braking torque. Teb depends on 1. DC source. 2. Rotor resistance 3. Speed

Consider rotor is running at syn speed Ns Stator is excited by DC The relative speed between stator field and rotor is Ns. Slip = (Relative speed Ns)/Ns = 1 This is equivalent to IM with a rotor at STANDSTILL Now consider, rotor is at rest and stator is excited by DC Stationary flux induces no rotor emf This is equivalent to IM with a rotor RUNNING at Syn speed Conclusions 1. Rotor at syn speed with DC dynamic braking is similar to rotor at rest during IM operation 2. Rotor at rest with DC dynamic braking is similar to rotor running at syn speed during IM operation

Circuit Diagram DC

AC

R1 Stator

Rotor

R1 is connected to limit stator current Additional rotor resistance is also connected to limit the current and to obtain braking characteristics

Circuit Diagram AC

Rectifier R1 Stator Rotor

Transformer

Under normal operating condition Rotor speed w r t stator field under DC dynamic braking is Ns(1-s) = NsS In the equivalent ckt diagram, replace s by S In phasor diadram jx2 I1 I2 also replace s by S V1

IΦ VDC

jXm

r2/S

I1

I1’ I0

x1 = 0, and no stator core loss

I2r2

SE2 I2 jI2Sx2

The dynamic braking torque is Ted = Te

3

ωs

I2

2

r2 S

The T-s ch is similar to IM but with slip scale reversed R2’ < R2’’< R2’’’ TL

A

Ns 0 r2 Ted R2’ R2’’ R2’’’ Ted increases with increase in rotor circuit resistance 0 Speed 1 Slip

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 = 3 I m N 2

MMF produced by single ph due to DC = IDC N The resultant MMF produced due to DC

IDC N

60

IDC N √3IDC N

∴For equal MMF due to AC and DC 3 I m N = 3IDC N 2 3 I1 N IDC = 2

MMFDC =

3IDC N

AC dynamic braking in nothing but SEIG operation A bank of capacitors is connected across three phases of stator wdg. IG receives AC excitation from bank of capacitor The generated electrical energy is dissipated as heat in rotor circuit AC Due to high cost of capacitor, this method is not used in practice. C Stator Rotor

C

Advantages of Dynamic Braking 1. Smooth stop

2. Less rotor ckt loss C 3. No tendency to reverse Disadvantage: Less quick than plugging

UNIT-5 Single phase Induction Motor & Special Machines Presented by C.GOKUL AP/EEE

Single phase Induction Motor

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

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 motors Capacitor-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 Advantages The 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.

Disadvantages Since capacitor start motors have low starting torque they cannot be used in applications with severe starting conditions.

Application of single phase capacitorrun 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 permanentsplit 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 startand-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 Advantages Simple in construction Tough surface Reliable and cheap

Disadvantages Low starting torque Very little overload capacity Low efficiency (5% for tiny sizes – 35% for higher ratings)

Applications of single phase shadedpole 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 available Here, 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

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

Variable Reluctance Stepper Motor 

2-phase-ON mode

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

Variable Reluctance Stepper Motor 





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

Variable Reluctance Stepper Motor      

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

Variable Reluctance Stepper Motor 

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

Variable Reluctance Stepper Motor

Variable Reluctance Stepper Motor  

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

Variable Reluctance Stepper Motor 



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

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

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

Permanent Magnet Stepper Motor  



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

Permanent Magnet Stepper Motor 

Consider positive current in phase A

Permanent Magnet Stepper Motor

Permanent Magnet Stepper Motor    

Advantages Permanent magnets require no external exciting current – low power loss High inertia Develops more torque than VR motor

Permanent Magnet Stepper 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

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

Hybrid Stepper Motor 

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

Hybrid Stepper Motor  

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 case It 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 chromesteel 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

AC series motor – Modifications Implementation   

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

AC series motor – Modifications Implementation 



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

Repulsion Motor - Operation 

To make it clear

Repulsion Motor - Operation 

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

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

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 Motor – Overcoming the disadvantages  

 

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 Motor – Overcoming the disadvantages      

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

Repulsion Motor – Overcoming the disadvantages 

 

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 fullload 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

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

Linear Induction Motor   

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

Linear Induction Motor 

 

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

Linear Induction Motor 

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

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  

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These motors are either separately-excited dc motors or permanentmagnet 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 

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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 phasedisplaced 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 Set point

+

Reference Interface input + E(s) Circuit

Intel micrcontroller Ts Digital z Controller o E*(s) h

Maglev Front Panel Actual Ball position Y(s)

Interface Magnetic Circuit U(s)Levitation System

References • Electrical Machines-II by S. B. Sivasubramaniyan -MSEC, Chennai • http://yourelectrichome.blogspot.in/ • http://www.electricaleasy.com/p/electricalmachines.html • www.scribd.com • 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

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