Tài liệu Induction Machines pptx

Stator of induction machines, together with its three-phase winding, completelycorresponds to the stator of synchronous machines.. Semi-closed slotsare usually selected for power ratings

Fig.1 – Cross-section of a three-phase induction machine.

Before proceeding further into discussion of operating principles and analytical theory

of induction machines, let us here briefly review the main constructional features of inductionmachines Stator of induction machines, together with its three-phase winding, completelycorresponds to the stator of synchronous machines This means that the stator core isassembled of laminated iron sheets An appropriate number of iron sheets are put together,thus forming stator core of the necessary length Laminated iron sheets are insulated in order

to reduce eddy-current losses in the iron core Such a design of iron core is always utilised forthose parts of electric machine where flux density and magnetic flux is time varying The sheetsare mutually isolated in order to prevent formation of a circuit for eddy-current flow fromsheet to sheet The iron core material is silicon alloyed Addition of silicon reduces hysteresis

losses in the iron core Simultaneously, electric resistivity of the material is increased, thusgiving rise to a substantial decrease in eddy-current losses as well Windings of inductionmachine are placed into slots, which may be of open shape, or semi-closed Semi-closed slotsare usually selected for power ratings up to 200 kW; above 200 kW slots are open.

Rotor core in induction machines is of laminated structure as well, because in rotorwindings flows AC current, giving rise to time-varying flux density in the rotor (in synchronousmachines rotor frequency is zero and rotor can be manufactured using a solid piece offerromagnetic material) Rotor winding is placed into rotor slots in one of the two differentways Subdivision of induction machines into two categories is normally done in conjunctionwith the way in which the rotor winding is formed Rotor winding may be wound three-phasewinding and such a machine is called wound-rotor induction machine or slip ring inductionmachine The latter name stems from the fact that three terminals of the rotor phases are inwound-rotor machine brought out of the machine and connected to three slip rings (one foreach phase), while the remaining three ends of the phase windings are connected into starneutral point and they remain inside the machine Wound-rotor induction machine is illustrated

in Fig 2 The purpose of slip rings is beyond the scope of interest here Slip rings aremechanically fixed to rotor and they rotate together with rotor Three carbon brushes aremounted on the stator (one for each phase) and brushes slip along the rings as they rotate, thusestablishing an electric contact between the stationary world and the rotating rotor An electricapproach to rotor winding (which rotates together with rotor) is enabled in this way In otherwords, electric energy can be either brought or taken away from rotor winding during machineoperation through this assembly, which consists of rotating slip rings and fixed brushes.Brushes are connected with leads to rotor winding terminals in the terminal box of themachine Slip rings can be short-circuited and then rotor winding becomes a three-phase short-circuited winding When the slip rings are short-circuited, brushes are raised and detached fromslip rings This is the normal operating state of a slip-ring induction machine

The second type of induction machines are so-called squirrel-cage induction machines.Rotor winding is in this case cast into slots and it is formed of solid aluminium or copper bars.The both ends of bars are electrically connected through end-rings The winding manufactured

in this way resembles a squirrel cage and this explains the origin of the name The winding isshown in Fig 3 Note that rotor winding is in this case always short-circuited and there is nopossibility of electrical approach to the winding In other words, electric energy cannot beeither brought or taken away from the rotor winding A winding formed in this way is

essentially an n-phase winding, where the number of phases n equals number of bars However, such an n-phase winding can be always substituted with an equivalent three-phase winding for

all analytical considerations

As already stated at the beginning of this section, there are rotor currents in the rotorwinding whose frequency is ωr We have just seen that in a squirrel-cage induction machinethere is no electrical access to the rotor winding Similarly, the normal operating regime of aslip-ring induction machine is with rotor winding short-circuited; thus no electrical access ispossible to the rotor winding and the question which arises is how do we get currents in rotorwindings when we cannot approach the winding and connect an appropriate electric source

2 OPERATING PRINCIPLES OF INDUCTION MACHINES

Consider an induction machine with a three phase winding on the stator and anequivalent three phase short-circuited winding on the rotor Let the stator winding beconnected to the utility supply which provides three phase balanced system of AC voltages.These voltages will cause appropriate three phase currents to flow through stator the winding

The currents will give rise to formation of Tesla’s revolving field in the air gap of the machine.

Given the supply frequency of stator voltages and currents f s in [Hz], the revolving field willrotate in space (in the cross section of the machine) at the angular frequency of ωs =2πf s Asboth stator and rotor are initially stationary, the revolving field cuts conductors of both statorand rotor windings, causing induced electromotive forces in the windings to occur

slip ring

brush winding

terminal A

B

C

Three-phase winding

Slip rings

Brushes Terminals

Shaft

Fig 2 – Schematic representation of the rotor of a slip ring induction machine and the

physical appearance.

In stator phase a winding a counter electromotive force is induced and it balances the applied

voltage, the difference in their rms value being caused by voltage drop on the windingresistance and leakage reactance and being equal to a couple of percents of the applied voltagerms value An electromotive force is induced in the rotor winding as well and its direction isshown in Fig 4 As the rotor winding is short-circuited, the induced electromotive force will

cause a current in the rotor conductors I r, whose real component has the same direction as

induced emf As the conductor, which carries current I r, is in the magnetic field, a magnetic

force F will be created This force will cause rotor to start rotating in the direction of stator

revolving field rotation The same happens in all the rotor conductors and sum of all individualmultiples of rotor radius and force gives overall electromagnetic torque in the machine.Summarising, when stator winding of a three-phase induction machine is connected to themains, electromagnetic induction causes currents in rotor windings and a torque is createdwhich pulls rotor into rotation in the direction of rotation of the stator revolving field Thisimplies that transfer of electric energy from stator to rotor is realised exclusively by

electromagnetic induction; therefore asynchronous machines are called induction machines.

Rotor winding bars End ring

Fig 3 – Squirrel-cage winding of a squirrel cage induction machine.

S

Fig 4 – Creation of an electromagnetic torque in an induction machine.

Rotor can never reach synchronous speed of rotation Rotation of rotor at synchronousspeed implies that rotor rotates synchronously with revolving field In that case there is norelative motion between stator revolving field and rotor and no electromotive force can beinduced in the rotor windings Consequently, no current can flow in the rotor winding if the

speed is synchronous and no electromagnetic force can be generated Therefore at synchronousspeed the developed torque in an induction machine equals zero As certain amount of torque

is always necessary in a machine that operates as a motor in order to cover mechanical losses,induction motor has to operate with certain amount of developed torque even when it is notloaded at the shaft Thus the rotor in motoring regime can never attain synchronous speed, i.e

it can never catch with the revolving field When the motor is unloaded, it runs under no-loadconditions and the amount of torque that is needed is determined with mechanical losses(windage losses and friction losses in bearings) The torque that describes mechanical losses issmall, thus indicating that the induction motor will have the highest possible speed when it runsunloaded; this is so-called no-load speed and it is only slightly smaller from synchronous speed

Let us summarise the above given explanations: connection of three-phase statorwinding of an induction machine at standstill to a voltage source causes current flow in statorwindings; these currents give rise to production of revolving field; revolving filed cutsconductors of both stator and rotor windings; emf is induced in stator and it provides voltagebalance to supply voltage; emf is induced in rotor as well and it causes current flow throughshort-circuited rotor winding; an electromagnetic force is created which acts on every rotorconductor, leading to the creation of the electromagnetic torque which pulls rotor intorotation; the direction of rotation is the same as the direction of rotation of stator revolvingfield; when a steady-state is established, rotor rotates with angular velocity equal to

Windings are by the virtue of their construction of resistive-inductive nature Reactivepower has to be provided for magnetisation of iron cores and air gap between stator and rotor.The question is how this reactive power is provided in induction machines The machine doesnot contain any capacitances that could produce reactive energy The only electricalconnection with outside world is the connection of the stator winding to the supply, as therotor winding is short-circuited This means that there is no source of reactive power availableinside an induction machine Therefore induction machine has always to absorb reactive energyfrom the supply Under all the possible operating conditions induction machine will act as areactive energy consumer As there is no rotor winding connected to another electric source,

as is the case in synchronous machines, there is no way of exciting the induction machine in amanner similar to synchronous machines This is one of the main reasons why inductionmachine is mainly utilised in motoring regime, while synchronous machine is used forgeneration purposes When an induction machine is applied as a generator, reactive power has

to be either taken from the power system or to be provided by a static VAr compensator (e.g.,capacitor bank)

As already emphasised, during motoring induction machine has to rotate slower thanthe revolving field, even under no-load conditions The angular velocity of the rotor is givenwith ω =ωs −ωr Revolving fields of stator and rotor rotate with angular velocity ωs The

difference between rotor speed and synchronous speed is characterised with the so-called slip.

The slip is expressed either as a percentage value of the synchronous speed or as a per unitnon-dimensional quantity It is usually calculated out of the speeds given in [rpm] in thefollowing way:

[%]

100or

[p.u]

s s s

s

n

n n s n

n

n

(1)where:

n s - synchronous mechanical speed, which is a function of the number of magnetic pole

pairs P and which is correlated with synchronous electrical speed 60f s as

n - asynchronous mechanical speed of rotation of induction machine shaft

Note that definition of the slip and the values are the same regardless of whether speeds in[rpm] or angular speeds in [rad/s] are used

Slip during normal operation of induction machines is in the range 10% to 2% forinduction machines with power ratings in the range 1 kW 100 kW The value of the slip that

corresponds to the rated operating conditions, when speed is n n , will be denoted as s n Index n

will in general always define the rated (nominal) operating condition of the machine

Let us now investigate correlation between stator and rotor frequencies with respect tonewly introduced notion of slip From slip definition of (1) it follows that

at the

Since

1260by

divided

(3)

it follows that the rotor frequency is determined with

s r s

Solution:

[Hz]

5 1 50 03 0

[rpm]

1455 1500 ) 03 0 1 ( 1

[rpm]

1500 2 / 50 60 60

f

x -s)n

(

n

x P

f n

s r

s

s s

[ [%]

025 0 3600 / ) 3150 3600 ( / ) ( ]

p.u.

[

[rpm]

3600 1 / 60 60 60

n n n s

x P

f n

s s

s s

According to (4), frequency of the current in rotor is slip times frequency of statorcurrents For 50 Hz stator frequency and operating slips of 10% to 2% in an induction machinerotor frequency is only 5….1 Hz Consequently, as the losses in the iron core are proportional

to frequency and frequency squared, it follows that rotor iron losses are going to be negligiblysmall and that the major part of the iron loss will take place in stator The total iron loss in aninduction machine is for this reason always assumed to take place in stator only

According to the slip definition, equation (1), slip is a variable determined with thespeed of rotation This implies that frequency of rotor currents is, according to (4), a variable

as well, proportional to the slip Characteristic slip values in motoring operation are:

0 < n < n0 rotor rotates, machine is loaded 1 > s > s0

n = n0< ns no-load, machine is unloaded s = s0

n = nn< n0 rotor rotates, rated load s = sn

Normal operating range of induction machines in steady-states is in the speed range betweenrated speed and no-load speed, the actual operating speed being dependent on the load torquethat the motor is driving

Suppose now that a source of mechanical energy is connected to the induction machineshaft and that the mechanical power provided by mechanical source is exactly equal to thepower which describes mechanical losses (i.e mechanical source provides torque to overcomemechanical loss torque) Then the speed of rotation will become equal to synchronous, as themechanical loss torque is equated by torque of the prime mover Simultaneously the inductionmachine torque will become equal to zero Therefore at synchronous speed

n s

It can be shown that a balanced three-phase induction machine fed from symmetricalsinusoidal three-phase supply can be treated in terms of per-phase equivalent circuit in steadystate However, such a derivation is pretty involved and time consuming As only steady statesunder symmetrical supply conditions are of interest here, complex representatives of ACsinusoidal quantities may be used (phasors) Furthermore, for the purpose of steady-stateanalysis of a balanced induction machine fed from symmetrical source, the whole analysis can

be performed by utilising per phase representation with complex phasors Such an approach isutilised in what follows

Let stator winding be connected to mains, which provide symmetrical three-phasevoltages and let rotor be at standstill, so that rotor speed is zero Frequency of stator voltages

and currents is f s An electromotive force will be induced in rotor winding that will causecurrent to circulate around the rotor winding As the rotor is at standstill, slip equals one andthe frequency in rotor winding equals stator frequency When the rotor is at standstill,difference between the speed of the rotating field and the rotor speed is of maximum value andequals synchronous speed This speed difference determines the induced emf, since the emf isdirectly proportional to the speed at which the conductors are cut by the field (i.e to thedifference between the synchronous speed and the rotor speed) Once when rotor rotates atcertain speed, the speed at which conductors are cut by the rotating field will be determinedwith ωr =ωs −ω =2πsf s and will be smaller than at standstill

When the rotor is at standstill let the induced electromotive force in one rotor phase E

is identified with index rl Its existence will cause current flow and rotor currents produce

corresponding revolving field and flux One part of the flux dissipates around the rotor winding(leakage flux), while major part links with stator windings contributing to the mutual flux.Current flow through rotor winding, caused by induced emf, is opposed by the resistance ofthe rotor winding and leakage reactance (which describes leakage flux) The value of the rotorleakage reactance is again frequency dependent At standstill rotor frequency, rotor leakagereactance and modulus of rotor current are

2 2

22

rl r

rl

rl

r s r

Suppose now that the rotor starts rotating, so that slip becomes smaller than 1 sincespeed is greater than one According to the fundamental expression for induced electromotiveforce due to the relative movement of a conductor with respect to flux density, induced emf isproportional to the relative speed of conductor with respect to flux density As the rotorrotates with certain speed, while revolving fields rotate with synchronous speed, relative speed

of the revolving field with respect to the rotor conductors is equal to the rotor angular

frequency This means that induced electromotive force at slip s is proportional to frequency of

rotor currents Consequently, at any other speed different from zero,

2 2 2 2

2

11

r r

rl

r

rl r

rl r

R

E

I

X s R

sE X

E

s sE

E

γ

γ γ

r

rl

rl r

rl r

r

r

I jX I

s

R

E

E s I jsX I

R

E

γ

γ+

=

−

−

=+

=

−

(8)

Equation (8) enables construction of the rotor per-phase equivalent circuit, shown in Fig 6

Let us consider now voltage balance for one stator phase winding Stator phasewinding is characterised with resistance and stator leakage reactance Note that for statorrotating field always cuts the conductors at the same, synchronous speed Hence the frequency

of the stator is constant (50 Hz) and the induced emf is proportional to this fixed frequencyregardless of the speed of rotation of the rotor The induced emf exists in each stator phase and

it holds balance to the applied stator voltage Following the same approach as for the rotorphase, one can immediately write the phasor voltage equation for one stator phase as:

s s s s

Fig 7 – Stator per-phase equivalent circuit.

By combining Figures 6 and 7, resulting complete equivalent circuit can be constructed

It is shown in Fig 8 and is described with the following two voltage phasor equations:

s s s s

R

r rl r

Fig 8 – Induction machine per-phase equivalent circuit.

Two circuits shown in Fig 8 apply to two different voltage levels, since the inducedemf in stator is in general different from the induced emf in rotor It is therefore not possible todirectly connect them In order to be able to put the two circuits together, it is necessary toapply transformer theory, which means that rotor voltage needs to be referred to stator voltagelevel (as the secondary is referred to primary using the transformation ratio in transformers).Correlation between the stator and rotor induced emf is established at standstill through thetransformation ratiom Transformation ratio is defined as

rl s

r

rl

r rl r

r

rl

I jmX I

=

−

•+

2

r rl r

r

r

r rl r

r

rl

X m X R

m

R

I jX I

s

R

E

γ γ

γ

2 2

''

'''

The left-hand side of (16) is now, by definition in (12), equal to the stator induced emf It istherefore possible to connect the two circuits into a single circuit, shown in Fig 9, which isdescribed with the following two voltage equations:

( r' rl') r'

s

s s s s

s

I jX s

R

E

E I jX

=

−

++

Fig 9 – Connection of the stator and rotor per-phase equivalent circuit into a single electric

circuit after referring rotor to stator.

Note that the phasors are identified in all the equations with a bar over the symbol, while inFigs 6-9 (and subsequent figures as well) phasors are denoted with a bold symbol

What remains to be done in Fig 9 is to express the induced emf in terms of anappropriate impedance This can be done in the same manner as in transformer theory, byutilising the notions of the magnetising current and magnetising reactance Phasor sum of thestator and referred rotor current is defined as the magnetising current, and induced emf isexpressed as a voltage across the magnetising reactance caused by the flow of the magnetisingcurrent:

to be taken into consideration, then the equivalent circuit of Fig 10 becomes as in Fig 11

by means of experiments (described in the next section) are anyway rotor parameters referred

to stator (i.e as seen from the stator side)

Calculate slip, power factor, stator current and input power of the motor when itoperates under rated conditions

Solution:

Note that since the rated speed is 1746 rpm and the motor is four-pole, the operating frequency must

be 60 Hz, since rated slip has to be positive and of the order of a couple of percents.

03 0 1800 / ) 1746 1800 ( / ) (

[rpm]

1746

[rpm]

1800 2 / 60 60 / 60

n

s s

n n n

s

n

x P f n

The motor is star connected Hence the phase voltage is 220/ √3 = 127 V.

In order to find the stator current, it is necessary to solve the equivalent circuit of Fig 10, in which the magnetising reactance is included From the circuit of Fig 10 one has:

Ω +

= +

+ +

= + + +

=

Ω

=

Ω +

= +

=

) 87 1 525 4 ( ) 52 1 135 4 ( ) 35 0 39 0 ( 16

) 35 0 67 4 ( ' '

j j

j Z

Z

Z Z jX R

Z

j Z

j jX

s R

Z

r m

r m s s in

m

rl n r r

γ γ

The modulus and the phase of the input impedance are

o

48 22 ) 897 4 / 525 4 ( cos } / ] {Re[

cos

897 4 87 1 525 4

1 1

2 2

=

−

−

in in n

in

Z Z Z

φ

Stator rated current and power factor are therefore

924 0 cos

A 26 9 4 / 127 /

) (

X14

.0R0.39

Rs = Ω 'r = Ω γs = 'γr = Ω

The motor is four-pole, three-phase, with star connected stator winding, ratedfrequency is 60 Hz and rated voltage is 220 V Rated speed is 1746 rpm Mechanicaland iron loss may be neglected Magnetising reactance may be regarded as of infinitevalue

Calculate slip, power factor, stator current and input power of the motor when itoperates under rated conditions

Solution:

Note that all the data in this example are the same as in the previous one The only difference is that the magnetising reactance is now omitted from the equivalent circuit Such an approximation enables much simpler calculations However, it is unrealistic for normal operating slip values, since it yields

an unrealistically high value of the power factor Rated slip value is calculated as in the previous example,

03 0 1800 / ) 1746 1800 ( / ) (

[rpm]

1746

[rpm]

1800 2 / 60 60 / 60

n

n

s s

n n n

s

n

x P f n

and the rated phase voltage is again 127 V Impedance calculation is however now much simpler,

W 9398 99 0 9 24 127 3

99 0 108 5 / 06 5 cos

A 9 24 108 5 / 127

108 5 7 0 06 5

) 7 0 06 5 ( ) 35 0 67 4 ( ) 35 0 39 0 (

) 35 0 67 4 ( ' '

2 2

=

Ω +

= +

+ +

= + +

=

Ω +

= +

=

x x x P

I

Z

j j

j Z

jX R

Z

j jX

s R

rl n r r

φ

γ γ

Comparing the results obtained with and without the magnetising branch, one can see that the error in the stator rms current is rather small (26 A against 24.9) Similar conclusion applies to the real input power (9131 W against 9398 W) However, the error in the power factor is significant (0.924 against 0.99) since when the magnetising branch is neglected reactive power taken by the machine is almost zero (only reactive power for the leakage reactances exists now, in contrast to the real case when most

of the reactive power appears across the magnetising reactance).

No-load and mechanical short-circuit (locked-rotor, rotor at standstill) are twoimportant regimes of an induction machine which occur during normal operation and which areperformed as standard tests on induction machines as well When performed as tests, these twotests enable calculation of the parameters of the equivalent circuit, determination of mechanicallosses and calculation of iron core losses Here no-load state and rotor at standstill conditionare discussed only as regimes that appear in normal operation of the motor