“mini project”

In partial fulfillment of the requirements

For The Degree of

Bachelor of Technology

Electrical & Electronics Engg.

MINI PROJECT TOPIC

“SLIP POWER RECOVERY of INDUCTION MOTOR”

Submitted to

Pranveer Singh Institute of Technology

Bhauti, NH-2, Kanpur

U.P. Technical University

Under the supervision of :-

Mr. Sanjay Kumar

(Asst. Professor Electrical Department)

(PSIT, Kanpur)

Name of the Scholars :-

Amit kumar Gautam 0816421010

Pulkit Gupta 0816421034

Vaibhav Diwedi 0816421049

SachinPrakash Bhatt 0816421408

November, 2011

CERTIFICATE

This is to certify that Project Report entitled “SLIP POWER OF INDUCTION

MOTOR” which is submitted by Amit kumar Gautam (0816421010), Pulkit Gupta

(0816421034), Vaibhav Diwedi (0816421049), Sachin Prakash Bhatt (0816421408) in

the partial fulfillment of the requirement for the award of degree B.Tech. in department of

ELECTRICAL & ELECTRONICS ENGG. Of U.P. Technical University, is record of the

candidates own work carried out by him under our supervision. The matter embodied in this

thesis is original and has been submitted for the award of any other degree.

Date:- SUPERVISOR

Mr. Sanjay Kumar

( Asst. Professor Electrical Department)

(PSIT, Kanpur)

DECLARATION

I hereby certify that this submission is own work and that, to the best of my

knowledge and belief, it contains no material previously published or written by another

person nor material which to a substantial extend has been accepted for the award of

Pranveer Singh Institute of Technology Kanpur at our organization to fulfill the

requirements for the award of degree of B.Tech (Electrical & Electronics Engineering).

During her tenure with us we found her sincere and hardworking. We wish her a great

success in the future.

Date:- Signature of the Incharge:-

Mr. Brijesh Kumar Dubey

(Sr. Lecturer)

Name of the Scholars :-

Amit kumar Gautam 0816421010

Pulkit Gupta 0816421034

Vaibhav Diwedi 0816421049

Sachin Prakash Bhatt 0816421408

ACKNOWLEDGEMENT

It give us a great sense of pleasure to present the report of the B.Tech Project

undertaken during B.Tech. Final Year. We own special debt of gratitude to Asst. Professor

Mr. Brijesh Kumar Dubey , Department of Electrical & Electronics Engineering,

Pranveer Singh Institute of Technology, Kanpur for his constant support and guidance

throughout the course of our work. His sincerity, thoroughness and perseverance have been a

constant source of inspiration for us. It is only his cognizant efforts that our endeavors have

seen light of day.

We also take the opportunity to acknowledge the Asst. Professor Mr. Sanjay Kumar,

(Asst. Professor Electrical & Electronics Engineering, Pranveer Singh Institute of

Technology, Kanpur for his full support and assistance during the development of the

project.

We also do not like to miss opportunity to acknowledge the contribution of all faculty

members of the department for their kind assistance and cooperation during the development

of our project. Last but not the least, we acknowledge our friends for their contribution in the

completion of the project.

Name of the Scholars :-

Amit kumar Gautam 0816421010

Pulkit Gupta 0816421034

Vaibhav Diwedi 0816421049

SachinPrakash Bhatt 0816421408

Date:-

ABSTRACT-

Induction motor drive, considering its significantly industrial applications, has great

importance. Squirrel cage IM are generally controlled by varying voltage and frequency fed

to the motor. For high power drive, because of use of forced commutations converters,

squirrel cage IM have less application. In this case and particularly when load torque is

proportional to high order of speed, wound rotor IM is used. In these motors speed is

controlled by energy absorption from rotor and returning it to network.

Return of slip energy to network improve efficiency, but Small power factor and harmonic

injection rest as a predominant problem in this drive system. In this paper slip energy

recovery method will be studied by dynamic modelling . To study Slip energy recovery

system and effect of the method on Harmonics, power factor, torque-speed characteristics

are determined.

INTRODUCTION-

Through the well known advantage of induction motor, these motor are widely used in

different industrial. The squirrel cage type of IM is generally controlled by controlling stator

voltage and frequency. In high power drive we had to use converter with force

commutations, practically. This increases the cost and the complexity of the control system.

For this reason, in high power requirement, squirrel cage IM motor drive have less

application. In such case, wound rotor IM is suggested. In these motors we can vary the

speed of motor by adding external resistor to rotor. In this method the slip energy

dissipated, into heat, in external resistance of motor can be utilised . If someway, we can

return the slip energy into network, we can increase the efficiency. Slip energy recovery

induction motor drive is one of the suggested methods for this purpose. In this method, by

varying firing angle of inverter, slip energy is absorbed from rotor and is returned to AC

network. Hence by varying firing angle we can return slip energy to network and motor

speed is controlled .Converter that utilized in this system has natural commutation (line-

commutated inverter) that makes it easier and less cost than converter with forced

commutation. Main problem in this drive is low power factor (because of reactive power

consuming by power electronic converter), and inject current harmonic in to the AC

network [4,5]. Furthermore this harmonic increase torques ripple and motor temperature.

INDUCTION MOTOR-

INDUCTION

MOTOR

SLIP-POWERCONVERTER

MAIN POWER SUPPLY

Slip-power return to supply

slip-power from rotor ckt

Induction motor is called induction motor because it works on electromagnetic

induction principle . EMF induced in rotor ckt. due to induction of magnetic flux . Induction

motor runs with asynchronous speed i.e. it runs the speed lesser than synchronous speed.

So it is also called Asynchronous Motor .

The least expensive and most widely spread induction motor is the squirrel cage

motor. The wires along the rotor axis are connected by a metal ring at the ends resulting in

a short circuit. There is no current supply needed from outside the rotor to create a

magnetic field in the rotor. This is the reason why this motor is so robust and inexpensive.

The stator phases create a magnetic field in the air gap rotating at the speed of the stator

frequency (we). The changing field induces a current in the cage wires which then results in

the formation of a second magnetic field around the rotor wires. As a consequence of the

forces created by these two fields, the rotor starts rotating in the direction of the stator field

but at a slower speed (wr). If the rotor revolved at the same frequency as the stator then

the rotor field would be in phase with the stator field and no induction would be possible.

The difference between the stator and rotor frequency is called slip frequency .

BASIC CONSTRUCTION -

The AC induction motor comprises 2 electromagnetic parts:

Stationary part called the stator

Rotating part called the rotor, supported at each end on bearings

The stator and the rotor are each made up of:

An electric circuit, usually made of insulated copper or aluminium, to carry current

A magnetic circuit, usually made from laminated steel, to carry magnetic flux

THE STATOR –

The stator is the outer stationary part of the motor, which consists of:

The outer cylindrical frame of the motor, which is made either of welded sheet

steel, cast iron or cast aluminium alloy. This may include feet or a flange for

mounting.

The magnetic path, which comprises a set of slotted steel laminations pressed into

the cylindrical space inside the outer frame. The magnetic path is laminated to

reduce eddy currents, lower losses and lower heating.

A set of insulated electrical windings, which are placed inside the slots of the

laminated magnetic path. The cross-sectional area of these windings must be large

enough for the power rating of the motor. For a 3-phase motor, 3 sets of windings

are required, one for each phase. The three coils form three windings distributed

over several slots. These windings may be connected in star or delta and three

terminations are brought out.

Stator and Rotor laminations

THE ROTOR:-

This is the rotating part of the motor. As with the stator above, the rotor consists of

a set of slotted steel laminations pressed together in the form of a cylindrical magnetic path

and the electrical circuit. The electrical circuit of the rotor can be either:

Wound rotor type, which comprises 3 sets of insulated windings with connections

brought out to 3 slip rings mounted on the shaft. The external connections to the

rotating part are made via brushes onto the slip rings. Consequently, this type of

motor is often referred to as a slip ring motor.

SLIP RINGS ON ROTOR

Squirrel cage rotor type, which comprises a set of copper or aluminium bars

installed into the slots, which are connected to an end-ring at each end of the rotor.

The construction of these rotor windings resembles a ‘squirrel cage’. Aluminium

rotor bars are usually die-cast into the rotor slots, which results in a very rugged

construction. Even though the aluminium rotor bars are in direct contact with the

steel laminations, practically all the rotor current flows through the aluminium bars

and not in the laminations. These bars are then shorted by rings that are brazed on

to each of the rotor ends. Such a rotor is called squirrel cage rotor. This rotor

behaves like a short-circuited winding and hence the machine is able to perform

electromechanical energy conversion. This type of rotor is easy to manufacture, has

no sliding contacts and is very robust. It is this feature that makes induction machine

suitable for use even in hazardous environments and reliable operation is achieved.

The disadvantage of this type of rotor is that the motor behaviour cannot be altered

by connecting anything to the rotor — there are no rotor terminals

SQUIRREL CAGE ROTOR

BASIC PRINCIPLE OF INDUCTION MOTOR -

when three phase supply is given to stator of 3-phase induction motor then a

rotating magnetic field is generated which rotates with synchronous speed .

Magnitude of this rotating flux is 1.5 times of maximum flux .

This rotating flux cuts the rotor conductors .

According to electromagnetic principle an emf is induced on rotor side due to flux

cutting action on rotor conductor .

If rotor ckt. is completed then a current starts flow on rotor side due to this emf .

This current produces a torque on rotor conductor.

If rotor is free to rotate then rotor starts rotating .

Direction of rotation can be determined by Lenz law means that the direction of rotation is

such that it always opposes the rotation of stator rotating flux. so it will rotate in the

direction of rotating field for minimising the flux cutting action of rotor .

The balanced three-phase winding of the stator is supplied with a balanced three-phase

voltage. The current in the stator winding produces a rotating magnetic field, the

magnitude of which remains constant. The axis of the magnetic field rotates at a

synchronous speed (ns =2*f/p), a function of the supply frequency (f), and number of poles

(p) in the stator winding. The magnetic flux lines in the air gap cut both stator and rotor

(being stationary, as the motor speed is zero) conductors at the same speed. The emfs in

both stator and rotor conductors are induced at the same frequency, i.e. line or supply

frequency, with No. of poles for both stator and rotor windings (assuming wound one) being

same. The stator conductors are always stationary, with the frequency in the stator winding

being same as line frequency. As the rotor winding is short-circuited at the slip-rings, current

flows in the rotor windings. The electromagnetic torque in the motor is in the same

direction as that of the rotating magnetic field, due to the interaction between the rotating

flux produced in the air gap by the current in the stator winding, and the current in the rotor

winding. This is as per Lenz’s law, as the developed torque is in such direction that it will

oppose the cause, which results in the current flowing in the rotor winding. This is

irrespective of the rotor type used − cage or wound one, with the cage rotor, with the bars

short-circuited by two end-rings, is considered equivalent to a wound one The current in the

rotor bars interacts with the air-gap flux to develop the torque, irrespective of the no. of

poles for which the winding in the stator is designed. Thus, the cage rotor may be termed as

universal one. The induced emf and the current in the rotor are due to the relative velocity

between the rotor conductors and the rotating flux in the air-gap, which is maximum, when

the rotor is stationary (nr =0.0). As the rotor starts rotating in the same direction, as that of

the rotating magnetic field due to production of the torque as stated earlier, the relative

velocity decreases, along with lower values of induced emf and current in the rotor. If the

rotor speed is equal that of the rotating magnetic field, which is termed as synchronous

speed, and also in the same direction, the relative velocity is zero, which causes both the

induced emf and current in the rotor to be reduced to zero. Under this condition, torque will

not be produced. So, for production of positive (motoring) torque, the rotor speed must

always be lower than the synchronous speed. The rotor speed is never equal to the

synchronous speed in an IM. The rotor speed is determined by the mechanical load on the

shaft and the total rotor losses, mainly comprising of copper loss.

The difference between the synchronous speed and rotor speed, expressed as a ratio of

the synchronous speed, is termed as ‘slip’ in an IM. So, slip (s) in pu is -

S = (ns –nr)/ns

Where , ns and nr are synchronous and rotor speeds in rev/s.

EQUIVALENT CIRCUIT OF AN INDUCTION MOTOR-

We have already seen that the induction motor can be treated as generalized transformer. Transformer works on the principle of electromagnetic induction. The induction motor also works on the same principle. The energy transfer from stator to rotor of the induction motor takes place entirely with the help of a flux mutually linking the two. Thus stator acts as a primary while the rotor acts as a rotating secondary when induction motor is treated as a transformer.If E1 = Induced voltage in stator per phase E2 = Rotor induced e.m.f. per phase on standstill k = Rotor turns / Stator turnsthen k = E2/ E1

Thus if V1 is the supply voltage per phase to stator, it produces the flux which links with both stator and rotor. Due to self induction E1, is the induced e.m.f. in stator per phase while E2 is the induced e.m.f. in rotor due to mutual induction, at standstill. In running condition the induced e.m.f. in rotor becomes E2r which is s E2.Now E2r = Rotor induced e.m.f. in running condition per phase R2 = Rotor resistance per phase X2r = Rotor reactance per phase in running condition R1 = Stator resistance per phase X1 = Stator reactance per phase So induction motor can be represented as a transformer as shown in the Fig. 1.

Fig. 1 Induction motor as a transformer

When induction motor is on no load, it draws a current from the supply to produce the flux in air gap and to supply iron losses.1. Ic = Active component which supplies no load losses2. Im = Magnetizing component which sets up flux in core and air gap These two currents give us the elements of an exciting branch as, Ro = Representing no load losses = V1 /Ic

and Xo = Representing flux set up = V1/Im

Thus, Īo = Īc + Īm

The equivalent circuit of induction motor thus can be represented as shown in the Fig. 2.

Fig. 2 Basic equivalent circuit

The stator and rotor sides are shown separated by an air gap. I2r = Rotor current in running condition = E2r /Z2r = (s E2)/√(R2

2 +(s X2)2) It is important to note that as load on the motor changes, the motor speed changes. Thus slip changes. As slip changes the reactance X2r changes. Hence X2r = sX2 is shown variable.Representing of rotor impedance : It is shown that, I2r = (sE2)/√(R2

2 +(s X2)2) = E2 /√((R2/s)2 + X22)

So it can be assumed that equivalent rotor circuit in the running condition has fixed reactance X2, fixed voltage E2 but a variable resistance R2/s, as indicated in the above equation. Now R2/s = R2 + (R2/s) - R2

So the variable rotor resistance R2/s has two parts.1. Rotor resistance R2 itself which represents copper loss.2. R2(1 - s)/s which represents load resistance RL. So it is electrical equivalent of mechanical load on the motor. Thus the mechanical load on the motor is represented by the pure resistance of value R2(1 -s)/s. So rotor equivalent circuit can be shown as,

Fig. 3 Rotor equivalent circuit

Now let us obtain equivalent circuit referred to stator side. Equivalent circuit referred to stator : Transfer all the rotor parameters to stator, k = E2/E1 = Transformation ratio E2' = E2/ k The rotor current has its reflected component on the stator side which is I2r'. I2r' = k I2r = (k s E2 )/√(R2

2 +(s X2)2) X2' = X2/K2 = Reflected rotor reactance R2' = R2/K2 = Reflected rotor resistance RL' = RL/K2 = (R2/K2)(1-s / s) = R2' (1-s / s) Thus RL' is reflected mechanical load on stator. So equivalent circuit referred to stator can be shown as in the Fig. 4

Fig. 4 Equivalent circuit referred to stator

.

The resistance R2' (1 -s)/ s = RL' is fictitious resistance representing the mechanical load on the motor.1.1 Approximate Equivalent Circuit Similar to the transformer the equivalent circuit can be modified by shifting the exciting current (Ro and Xo) purely across the supply, to the left of R1 and X1. Due to this, we are neglecting the drop across R1 and X1 due to Io, which is very small. Hence the circuit is called approximate equivalent circuit. The circuit is shown in the Fig.5.

Fig. 5 Approximate equivalent circuit

Now the resistance R1 and R2' while reactance X1 and X2' can be combined. So we get, R1e = Equivalent resistance referred to stator = R1 + R2' X1e = Equivalent reactance referred to stator = X1 + X2' R1e = R1 + (R2/K2)and X1e = X1 + (X2/K2)While Ī1 = Īo + Ī2r' .........phasor diagramand Īo = Īc + Īm Thus the equivalent circuit can be shown in the Fig.6.

Fig. 6

Power Equations from Equivalent Circuit With reference to approximate equivalent circuit shown in the Fig. 6, we can write various power equations as, Pin = input power = 3 V1 I1 cos Φ where V1 = Stator voltage per phase I1 = Current drawn by stator per phase cos Φ = Power factor of stator Stator core loss = Im

2 Ro

Stator copper loss = 3 I12 Ro

where R1 = Stator resistance per phase P2 = Rotor input = (3 I2r'2

R2')/s Pc = Rotor copper loss = 3 I2r'2

R2'Thus Pc = s P2

Pm = Gross mechanical power developed

T = Torque developed

where N = Speed of motorBut N = Ns (1-s) =, so substituting in above

and I2r' = V1 /(( R1e + RL') + j X1e )where RL' = R2' (1-s)/s I2r' = V1/√(( R1e + RL')2

+ X1e2

)

all the above formula all the values per phase values.

Consider the approximate equivalent circuit as shown in the Fig.7 In this circuit, the exciting current Io is neglected hence the exciting no load branch is not shown.... I1 = I2r' The total impedance is given by, ZT = (R1e + RL')+ where RL' = R2' (1-s)/s I1 = V1 /√((R1e + RL')2 +(X1e)2) The power supplied to the load i.e. Pout per phase is, Per phase Pout = I1

2 RL' watts per phase... Total = 3 I1

2 RL'

To obtain maximum output power, differentiate the equation of total Pout with respect to variable RL' and equal to zero.

But Z1e = √(R1e2 +X1e

2) = Leakage impedance referred to stator... Z1e

2 = RL'2

Thus the mechanical load on the induction motor should be such that the equivalent load resistance referred to stator is equal to the total leakage impedance of motor referred to stator.Slip at maximum Pout  : This can be obtained as, RL' = Z1e = R2'(1-s)/s where RL' = R2/K2

... s Z1e = R2' - sR2'

... s(Z1e + R2') = R2'

       This is slip at maximum output.

Expression for maximum   P out  : Using the condition obtained in expression of total Pout , we can get maximum Pout.... (Pout)max = 3 I1

2 Z1e as RL' = Z1e

But R1e2 + X1e

2 = Z1e

Maximum Torque

In case of induction motor, the speed of the motor decreases with increase in load. Thus

the maximum power output is not obtained at a slip which corresponds to maximum

torque. In the previous section we have seen the condition for maximum power output. In

this section we will find the condition which gives maximum torque.

The expression for torque is given by,

The condition for maximum torque can be obtained from maximum power transfer

theorem. When I2r'2 R2'/s is maximum consider the approximate equivalent circuit of

induction motor as shown in The Fig. 8.

Fig. 8

The value of Ro is assumed to be negligible. Hence the circuit will be reduced as shown

below.

Fig. 9

The thevenin's equivalent circuit for the above network is shown in the Fig.10 across the

terminals x and y.

Fig. 10

The mechanical torque developed by rotor is maximum if there is maximum power

transfer to the resistor R2'/s. This takes place when R2'/s equals to impedance looking back

into the supply source.

This is the slip corresponding to the maximum torque. The maximum torque is given by,

Substituting,

From the above expression, it can be seen that the maximum torque is independent of

rotor resistance.

Synchronous Watt

The torque produced in the induction motor is given by,

Thus torque is directly proportional to the rotor input. By defining new unit of torque

which is synchronous watt we can write,

T = P2 synchronous-watts

If torque is given in synchronous-watts then it can be obtained in N-m as,

Key Point : Unit synchronous watt can be defined as the torque developed by the motor

such that the power input to the rotor across the air gap is 1 W while running at

synchronous speed .

Determination of circuit parameters -

In order to find values for the various elements of the equivalent circuit, tests must be

conducted on a particular machine, which is to be represented by the equivalent circuit. In

order to do this, we note the following.

1. When the machine is run on no-load, there is very little torque developed by it. In an ideal

case where there is no mechanical losses, there is no mechanical power developed at no-

load. Recalling the explanations in the section on torque production, the flow of current in

the rotor is indicative of the torque that is produced. If no torque is produced, one may

conclude that no current would be flowing in the rotor either. The rotor branch acts like an

open circuit. This conclusion may also be reached by reasoning that when there is no load,

an ideal machine will run up to its synchronous speed where the slip is zero resulting in an

infinite impedance in the rotor branch.

2. When the machine is prevented from rotation, and supply is given, the slip remains at

unity. The elements representing the magnetizing branch Rm & Xm are high impedances

much larger than R′r & X′lr in series. Thus, in the exact equivalent circuit of the induction

machine, the magnetizing branch may be neglected . From these considerations, we may

reduce the induction machine exact equivalent circuit .

These two observations and the reduced equivalent circuits are used as the

basis for the two most commonly used tests to find out the equivalent circuit parameters —

the blocked rotor test and no load test. They are also referred to as the short circuit test and

open circuit test respectively in conceptual analogy to the transformer .

NO-LOAD TEST -

The no-load test, like the open circuit test on a transformer, gives information about exciting

current and rotational losses. The test is performed by applying balanced rated voltage on

the stator windings at the rated frequency. The small power provided to the machine is due

to core losses, friction and winding loses. Machine will rotate at almost a synchronous

speed, which makes slip nearly zero. This test is represented with an equivalent circuit in

Figure shown.

Values measured during this test are current and it’s angle with respect to Known voltage.

From this we can calculate total power supplied to the machine.

BLOCKED ROTOR TEST -

The locked rotor test, like short circuit test on a transformer, provides the information about

leakage impedances and rotor resistance. Rotor is at the stand still, while low voltage is

applied to stator windings to circulate rated current. Measure the voltage and power to the

phase. Since there is no rotation slip, s=1 which gives us following equivalent circuit.

SLIP POWER RECOVERY –

The slip power recovery (SPR) drive is an external system connected to the rotor circuit in place of the external resistors .The SPR provides speed and torque control like the resistors but can also recover the power taken off the rotor and feed it back into the power system to avoid energy waste.

An SPR drive consists of two interconnected power converters . The rotor converter is connected to the three-phase rotor winding. The feedback power converter is connected to the power system, usually through a transformer that matches the output voltage of the converter to the power system. The regenerative or feedback power converter is controlled to modulate the amount of power put back into the power system, allowing control of the motor speed. All the rotor energy previously lost as heat in the rheostat is now saved, and for large motors, this amounts to significant cost savings.

Slip energy recovery system analysis-

Fig. shows an induction motor that is included Slip energy recovery system .

In this configuration inverter is linked directly to the network [1,6].Resistance Rd and inductance Ld are used as current filter to reduce current ripple that passes through

inverter. This filter prevents discontinuous current, hence harmonic cupper loss will be reduced in motor. Relationship between rectifier and inverter can be expressed as:

Vd=q V √2 sin (π /q) / n1π =1.35 SV/n1 (1)

VI=1.35 V│COS(α)│ (2)

If we ignore the voltage losses in resistance, we find:

Vd=VI 1.35 SV/n1=1.35V│COS(α)│ (3)

S= n1│COS(α)│ (4)

With assume n1=1 we can write:

S= │COS(α)│ (5)

Consequently rotor speed can be obtained as:

ωr= ωe(1-│COS(α)│) (6)

Hence by changing fire angles between π/2<α<π, the speed vary from nominal speed to zero. This drive can change speed in sub synchronous zone , π/2<α <π then 0 < ωr < ωe (7)

Regarding maximum fire angle for thyristor (minimum advance angle β), the maximum fire

angle is limited to 150 0.

0 < S < 0.866 then 0.134 ωe < ωr < ωe (8)

As the mentioned system changes speed from nominal to 0.134 nominal speed, so the drive operates in motor region and can’t work in regenerative breaking at sub synchronous. Neglecting power loss in rotor, it can be written:

Slip power = SPg =VI Id (9)

Pm =(1-s)Pg = Te ωr (10)

Te = VI ID/S ωs (11)

Then Te = 1.35V / n1ωs ID =K.ID (12)

This expression shows that motor torque is proportional to the current transferred from rectifier to inverter (ID). If load torque increases, current ID increases to balance load torque & motor torque. For constant fire angle (VI=const), in order to compensate (VI*ID), because of ID enhancement, VD increased. Consequently torque-speed characteristics of this drive are similar to series DC motor. It can be noted that current ID is independent of motor speed and only depends on load torque. Neglecting power loss in rotor, current is proportional to load torque (TL) hence:

ID = K1TL (13)

This equation shows that for constant load torque, changing fire angle causes motor speed change even

II. AC & DC Side equivalent circuitNeglecting resistance and semiconductor switches voltage drops, transferred power can be written as:

RdId

2 +VIID = RdId

2+1.35V ID│COS(α)│ (14)

In each phase we can write:

P′ =1/3 Id2Rd +1/3 (1.35)V ID │COS(α)│ (15)

Pg=Ir2 Rr + P′ +Pm (pg=slip energy) (16)

In three-phase rectifier we can write:

Ir = (2/3)1/2 ID (17)

Irf = (6)1/2 ID / π (18)

Iref is fundamental component of the rotor current.

Power loss in each phase is equal to:

Pcu= Ir

2 Rr + 1/3 ID 2Rd=

Ir

2Rr+1/3 (3/2 Ir)1/2 Rd=Ir2 (Rr+.5Rd) (19)

Resistance loss produced by Iref is:

Pcu Irf = Irf (Rr+.5Rd) (20)

Fig.2

Mechanical power in each phase is :

Pm = [transfer power +loss power by fundamental current

component] (1-s)/s

Pm=[ Irf 2(Rr+.5Rd) +1.35V │COS(α)│ π/ (6)1/2 Irf ] (1-s)/s

Pg=Irf

2 RX + Irf 2 RA /S & RX=( (π2 /9 1)(Rr+.5Rd))

RA=(Rr+.5Rd) +1.35V │COS(α)│ / 3Irf(6)1/2 (21)

The ac side equivalent circuit is illustrated in Fig. 2. To obtain dc side equivalent circuit, we can transfer AC-side circuit to DC-side. Rotor current is consist of fundamental and harmonic component:

Irf = (6)1/2Id / π (23)

Ir = (2/3)1/2 Id (24)

For transferring AC-side circuit to DC-side, it can be

written as:

3Ir 2 ( S RS ′ +Rr )= IdRe Re=2SRS′+8Rr (25)

Re, is equivalent resistance in rotor. Considering XLs′, XLs, we should note that this reactance cause voltage drop equal to VX′ (due to overlap voltage drop) :

V X′ =S(3XC Id) / π =3S/ π (XLs′+ XLr)Id (26)

As effect of this reactance can be considered as a voltage drop VX or impedance equal to ′3S/ π (XLs +XLr). Equivalent circuit is shown in Fig.3.′

Fig.3 DC side equivalent circuit

In this circuit, voltage drop on resistance and semiconductor switches is not included. Torque can be written as:Te = 3Irf2 RA/Sωs (27)Irf = SVS /Req (28)′Req =[(SRS +RX +RA/S)2 +(SXLs +SXLr)2]/2(29)′ ′ in steady state condition, torque is equal to: Te=3 SVS 2 RA/ ωs [(S RS +RX +(RA/S))2 + (S XLs +SXLr)2] (30)′ ′ ′This equation shows relationship between motor torque, slip & fire angle. In Fig.4, motor torque is drawn regarding changes in S & α.

Fig.4: steady state torque for various load torque and fire angle

REFERENCES

www.google.com

www.wikipedia.com

www.physicsform.com

www.findpdfdoc.com

www.answer.com

www.scribed.com

www.indiamart.com

www.electricalengineering.com

www.allaboutcricuits.com

www.electronics.com

Book references

"Induction Motors" (2011), Electric Motors Reference Center, Machine

Design, Penton Media, Inc.

Babbage, C. and Herschel, J.W.F. (1825) "Account of the repetition of M.

Arago's experiments on the magnetism manifested by various substances

during the act of rotation," Philosophical Transactions of the Royal Society

of London, vol. 115, pages 467-496.

Silvanus Phillips Thompson, Polyphase electric currents and alternate-

current motors (London, England: E. & F.N. Span, 1895),

Prodigal Genius: The Life of Nikola Tesla .

Galileo Ferraris, “Electromagnetic rotation with an alternating current,"

Electrican, Vol 36 [1885].

M.D. Singh, K.B. Khanchandani, Power Electronics, Second Edition, Tata

McGraw-Hill, New Delhi, 2007, pages 148-152.

Dr. P.S Bimbhra “power electronics” second Edition, Khanna Publication.

Dr. P.S Bimbhra “Electrical Machinery” second Edition, Khanna Publication.