In partial fulfillment of the requirements For The Degree of
Bachelor of Technology Electrical & Electronics Engg.
MINI PROJECT TOPIC
Submitted to Pranveer Singh Institute of Technology Bhauti, NH-2, Kanpur
Under the supervision of :Mr. Sanjay Kumar (Asst. Professor Electrical Department) (PSIT, Kanpur)
Name of the Scholars :Amit kumar Gautam
0816421010
Pulkit Gupta
0816421034 0816421034
Vaibhav Diwedi
0816421049 0816421049
SachinPrakash Bhatt
0816421408 0816421408
November, 2011
CERTIFICATE This is to certify that Project Report entitled
MOTOR” which is submitted by Amit
“SLIP POWER OF INDUCTION
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 0816421034
Vaibhav Diwedi
0816421049 0816421049
Sachin Prakash Bhatt
0816421408 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 Professor Electrical Electrical & Electronics Engineering, Technology, Kanpur
Pranveer Singh Institute Institute of
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 :-
Date:-
Amit kumar Gautam
0816421010 0816421010
Pulkit Gupta
0816421034 0816421034
Vaibhav Diwedi
0816421049 0816421049
SachinPrakash Bhatt
0816421408 0816421408
Induction motor drive, considering its significantly industrial applications, 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 characteristics are determined.
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. 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 (linecommutated 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 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 s tator 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 .
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
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 electromagnetic principle an emf is induced induced on rotor side due to flux cutting action on rotor conductor .
If rotor ckt. is completed then a current 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 stator winding produces a rotating magnetic 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 (n r =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 MOTORWe 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 turns
then
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 E 1, 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 E 2r which is s E 2. 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 losses 2. 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 = V 1 /Ic and
Xo = Representing flux set up = V 1/Im
Thus,
Ī o = Ī c Ī m
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 2
2
= E2r /Z2r = (s E 2)/√(R2 +(s X2) ) It is important to note that as load on the motor changes, the motor speed changes. Thus slip changes. As slip changes the reactance X 2r changes. Hence X2r = sX2 is shown variable. Representing of rotor impedance :
It is shown that,
2
2
2
2
I 2r = (sE2)/√(R2 +(s X2) ) = E2 /√((R2/s) + X2 )
So it can be assumed that equivalent rotor circuit in the running condition has fixed reactance X2, fixed voltage E 2 but a variable resistance R 2/s, as indicated in the above equation.
Now
R2/s = R2 + (R2/s) - R 2
So the variable rotor resistance R 2/s has two parts. 1. Rotor resistance R 2 itself which represents copper loss. 2. R 2(1 - s)/s which represents load resistance R L. 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 R 2(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 I 2r'. 2
2
I2r' = k I2r = (k s E 2 )/√(R2 +(s X2) ) 2
X2' = X2/K = Reflected rotor reactance 2
R2' = R2/K = Reflected rotor resistance 2
2
RL' = RL/K = (R2/K )(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 R 2' (1 -s)/ s = R L' 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 R 1 and X1. Due to this, we are neglecting the 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 circuit
Now the resistance R 1 and R2' while reactance X 1 and X2' can be combined. So we get, R1e = Equivalent resistance referred to stator = R 1 + R2' X1e = Equivalent reactance referred to stator = X 1 + X2' 2
R1e = R1 + (R2/K ) 2
and
X1e = X1 + (X2/K )
While
Ī 1 = Ī o Ī 2r 2r' Ī o = Ī c Ī m
and
.........phasor diagram
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 V 1 I1 cos Φ where V1 = Stator voltage per phase I1 = Current drawn by stator per phase
cos Φ = Power factor of stator 2
Stator core loss = Im Ro
2
Stator copper loss = 3 I 1 Ro where
R1 = Stator resistance per phase 2
P2 = Rotor input = (3 I 2r' R2')/s 2
Pc = Rotor copper loss = 3 I 2r' R2' Thus
Pc = s P2 Pm = Gross mechanical power developed
T = Torque developed
where
N = Speed of motor
But
N = Ns (1-s) =, so substituting in above
and
I2r' = V1 /(( R1e + RL') + j X 1e )
where
RL' = R2' (1-s)/s 2
2
I2r' = V1/√(( R1e + RL') + X1e ) 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 I o 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 2
2
I1 = V1 /√((R1e + RL') +(X1e) ) The power supplied to the load i.e. P out per phase is, 2
Per phase Pout = I1 RL' watts per phase .
..
2
Total = 3 I 1 RL'
To obtain maximum output power, differentiate the equation of total P out with respect to variable RL' and equal to zero.
2
2
Z1e = √(R1e +X1e ) = Leakage impedance referred to stator
But
2
.
2
Z1e = RL'
..
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 ..
.
s Z1e = R2' - sR2'
.
s(Z1e + R2') = R2'
..
where RL' = R2/K
2
This is slip at maximum output.
Expression for maximum Pout : Using the condition obtained in expression of total P out , we can get maximum P out. .
..
2
(Pout)max = 3 I1 Z1e
But
2
as RL' = Z1e
2
R1e + X1e = 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 2
theorem. When I2r' 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 R 2'/s. This takes place when R 2'/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 noload. Recalling the explanations in the section on torque pro duction, 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 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 param eters — 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 K nown 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 e quivalent 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 analysisFig. 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 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 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 circuit Neglecting 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 +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 c ircuit 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
"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, Thompson, Polyphase electric currents and alternate-current motors (London, England: E. & F.N. Span, 1895),
Prodigal Genius: The Life of Nikola Tesla. 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 “Electrical Machinery” second Edition, Khanna Publication.
