Presentation on theme: "4. Equivalent circuit of IM."— Presentation transcript:
14. Equivalent circuit of IM. Developing an equivalent circuit model is useful to study and predict theperformance characteristics of the induction machine with reasonableaccuracy.In this section a steady-state per-phase equivalent circuit will be derived.For convenience, consider a three-phase wound-rotor induction machineas shown in fig 3.1.Fig phase induction machines equivalent circuitmodel
2If currents flow in both stator and rotor windings, resultant rotating magnetic fields will be produced inthe air gap.This resultant air gap field will induce voltages in bothstator windings (at supply frequency f1) and rotorwindings (at slip frequency f2).It appears that the equivalent circuit may assume a formidentical to that of a transformer; therefore it can bedeveloped from the knowledge of a transformerequivalent circuit and from knowledge of variation ofrotor frequency with speed in IM.The only difference is on account of the fact that the secondary winding (rotor winding) of an induction motor rotates and mechanical power is developed.
3The stator circuit model As explained earlier, when a voltage V1 is applied to the stator terminals, the rotating flux generates counter emf in the 3-phase of stator windings. The applied voltage V1 has to overcome counter emf = - E1, V1` and stator leakage impedance drop,I1(r1 + jx1). Thus,V1 = per-phase terminal voltageR1 = per-phase stator windingresistanceX1 = per-phase stator leakagereactanceE1 = per-phase induced voltage inthe stator windingXm = per-phase stator magnetizingreactanceRc = per-phase stator core loss resistanceFig The stator circuit model.
4The stator current consists of two components, the load component, I2` (-I2) and the exciting component Ie.I2` counteracts the rotor mmf, I2N2.The exciting component Ie, whose function is to create the resultant air gap flux and to provide the core loss, friction and windage loss and small stator and rotor I2R losses I0 is usually 30% - 50% of rated current of IM.Thus, the exciting current can be resolved into two components.- magnetising component, Im lagging V1` by 900 and- core loss component Ic which is in phase with V1`.In the equivalent circuit, Ic and Im are accounted for by a shunt branch consisting of:-- core loss resistance Rc = V1`/Ic- magnetising reactance Xm = V1`/Im
5Xm = ωsLs = 2πf1Ls Ls =(Ns)2/RL Since there is air-gap between stator and rotor of induction motor, Xm (reactance on account of effect of magnetizing current) will be much less than transformer case. Thus, Im will be much greater than in IM.Xm = ωsLs = 2πf1LsLs =(Ns)2/RLФ,wbTransformerInduction motorF,ATFig. 4.3 Flux mmf relation in transformer and IM.Note:- The graph indicates that, Reluctance of IM is muchgreater than reluctance of transformers.
6Rotor circuit modelRegarding the actual physical facts existing in the rotor circuit, the per phase rotor current is:Where, SE2 – slip frequency emfI2 – slip frequency currentIf the right hand side is divided by slip ,we get:- (for further analysis)Fig. 4.4a Rotor equiv.circuit at slip s.E2 – per phase equi. Circuit at stator frequency.I2Fig. 4.4b
7To determine the complete equivalent circuit of the rotor, the line frequency emf E2, the line frequency current I2 and the speed dependant resistance r2/s must be referred to the stator primary by the effective turn's ratio as in the transformer.Where, N1` = N1 Kw1N2` = N2Kw2
8fig.4.5b Exact equivalent circuit of IM Fig.4.5a Rotor equivalent circuit referred to the statorWhere,fig.4.5b Exact equivalent circuit of IM
9The equivalent circuit of IM is almost similar to the transformer equivalent circuit. i.e. - when S = 1, the exact equivalent circuitbecomes the equivalent circuit of a short circuitedtransformer.- In case S = 0, (at synchronous speed), the circuitbecomes the equivalent circuit of an open circuited
10Analysis of the equivalent circuit of IM The equivalent circuit shown in Fig.3.4b is not convenient to use for the determination of the performance of the induction machine .As a result, several simplified versions have been proposed.Some of the commonly used versions of the equivalent circuit are discussed here.
11Approximate Equivalent Circuit If the voltage drop across R1 and X1 is small and the terminal voltage V1 does not appreciably differ from the induced voltage E1, the magnetizing branch (i.e. Rc and Xm), can be moved to the machine terminals as shown in the fig. 3.6 below. (this is done in transformer equivalent circuit parameters determination)fig. 4.6 approx. equiv. circuit
12This approximation of the equivalent circuit will considerably simplify computation, because the excitation current (Ie) and the load component (I'2) of the machine can be directly computed from the terminal voltage V1 by dividing it by the corresponding impedance.Note that if the induction machine is connected to a supply offixed voltage and frequency, the stator core loss is fixed.Core loss = (hyst. + eddy current) losseshyst. Loss ∞ voltage appliededdy loss ∞ ( f )2At no load, the machine will operate close to synchronous speed.Therefore, the rotor frequency f2 is very small and hence rotorcore loss is very small.At a lower speed f2 increases and so does the rotor core loss.The total core losses thus increase as the speed falls.On the other hand, at no load, frication and windage losses aremaximum and as speed falls these losses decreases.
13Therefore, if a machine operates from a constant voltage and constant-frequency source, the sum of core losses and friction and windage losses remains essentially constant at all operating speeds.These losses can thus be lumped together and termed the constant rotational losses of the induction machine.If the core loss is lumped with the windage and frication loss, Rc can be removed from the equivalent circuit, as shown in the fig. below.Fig. 4.7
14IEEE RECOMMENDED EQUIVALENT CIRCUIT In the induction machine , because of its air gap, the exciting current Ieis high of the order of 30 to 50 percent of the full-load current.The leakage Reactance X1 is also high .The IEEE recommends that, in such situation, the magnetizing reactance Xmcan not be moved to the machine terminals (as is done in transformers.), butbe retained at its appropriate place , as shown in the fig below.The resistance RC is however, omitted, and the core loss is lumpedwith the windage and friction losses.This equivalent circuit is to be preferred for situation in whichthe induced voltage E1 differs appreciably from the terminal voltage V1.ZfabFig. 4.8 a. Equivalent circuit of IM without core loss.
15Steady state performance parameters of IM, such as current, speed, torque, losses, etc. can be computed from the circuit of fig. 3.8.In this fig., Zf is the per phase impedance offered by the rotating air-gap field.Note that Zf includesZfaFig 4.8b
16Impedance Z1 as seen by stator voltage V1 is:- AndThe total power lost inThe power lost intherefore,
17Determining the Equivalent circuit model parameters of IM. The response of an IM to changes in load can be determined from the equivalent circuit of the motor.The parameters R1,X1, R2, X2, Xm can be determined by performing a series of tests on the IM. i.e.- No-load test- DC test for stator resistance- Blocked rotor test.These tests need precision because the resistances vary with temperature and the rotor resistance also varies with rotor frequency.
18Fig. 4.9 Circuit diagram for no-load and blocked-rotor Test The No-load testThe induction motor is made to run at no load at rated voltage and frequency.Per phase values of applied stator voltage Vnl, input current Inl and input power Pnl are recorded ( by two wattmeter method).Fig Circuit diagram for no-load and blocked-rotor TestFig IEEE RECOMMENDEDEQUIVALENT CIRCUIT
19Then, the no-load reactance Xnl seen from the stator terminals is:- Since slip Snl at no-load is very small, r2/s at no-load is very large as compared to Xm.Thus, the total impedance of parallel branches consisting of jXm and (r2/s + jX2) is almost equal to jXm.Then, the no-load reactance Xnl seen from the stator terminals is:-jXnl = jX1 + jXm = X1 – stator self reactance.Fig Equivalent circuit under no-load test.
20From no-load test readings we can get; ,,If X1 is known, Xm can be determined.During no-load condition, as stated earlier, the rotational losses ( friction, windage and core losses are usually assumed constant and can be obtained from the relation,Rotational losses or fixed losses are:- Stator and rotor Core losses and friction and windage losses.
21i.e., no-load test gives Xnl and Prot Thus,The no-load test measures the rotational losses and gives information about its magnetisation current.i.e., no-load test gives Xnl and ProtNote that, the only load during no-load test is friction and windage loss.The mechanical power (Pm = Pg - Prcul) is consumed by mechanical losses.
22Friction and windage loss Shaft power Stator I2R lossStator core lossRotor I2R lossRotor core lossFriction and windage lossPgRotor input powerShaft powerMech. power developedInput power(3IVcosθ)PMFig Power flow diagram of IM.
23DC test for stator resistance The rotor resistance r2 plays an extremely critical role in the operation of an IM. i.e.- It determines the shape of the speed torque curve.- It determines the speed at which the pull-out torque occurs.To determine r2, it is necessary to find r1 of the stator.- Thus, a DC voltage is applied to the stator windings of an IM.- The current in the stator is adjusted to the rated value & the voltage ismeasured.If the supply dc voltage is connected between the two phases, ( the 3rd phase being open) then, the dc current flows through the two phase windings and the required per phase stator resistance is:-
24Fig 4.13 Circuit for Dc test to determine stator resistance w+ARDCVS_TFig Circuit for Dc test to determine statorresistance
25Blocked rotor test During this test; - An Ac voltage is applied to the stator; and the currentis adjusted to the rated value.- The rotor is blocked so that it can not rotate.- The current, voltage and power are measured.Since the rotor is kept stationary, the slip = 1,and the rotor resistance r2/s is just equal to r2 which is quite a small value.Fig IEEE RECOMMENDED EQUIVALENT CIRCUIT
26Compared with Xm, r2 & X2 are small, so that the whole input current flows through them. The equivalent circuit under this condition looks a series combination of r1, X1, X2, r2.IbrFig.4.15 equivalent circuit during blocked rptor test
27Impedance of the IM during block rotor test is:- Where, Xbr is the reactance of stator & rotorIn practice, it is difficult to get separately stator and rotor reactance, and are usually taken from experimental data from tables.
28Empirical distribution of leakage reactance Xbr However, for wound rotor machines x1 is assumed to be equal to x2. i.e. x1 = x2= ½XbrFor squirrel cage induction machines, total leakage reactance Xbr (=x1+ x2) can be distributed between stator and rotor as per the following table:Empirical distribution of leakage reactance XbrClass of motorFraction of XbrX1X2Class A (normal Tst , normal Ist and low slip)Class B (normal Tst, low Ist and low slip)Class C (high Tst , low Ist and low slip)Class D (high Tst , low Ist and high slip)0.50.40.30.60.7For design classes, please refer Stephen J. Chapman, 2nd ed.
290.5 0.4 0.3 0.6 0.7 Class A (normal Tst , normal Ist and low slip) Class of motorFraction of XbrX1X2Class A (normal Tst , normal Ist and low slip)Class B (normal Tst, low Ist and low slip)Class C (high Tst , low Ist and low slip)Class D (high Tst , low Ist and high slip)0.50.40.30.60.7
30Thus, the block rotor test helps to determine equivalent circuit parameters r2, Xbr(X1+X2). If X1 & X2 are taken from tables, then Xm can be determined. i. e.; Xnl = X1 + XmIn general, by performing these three tests and using experimental values of X1 and X2 from tables, we can determine the equivalent circuit parameters of IM.
31Separation of friction and windage loss from the no-load test result The power input to the induction motor at no-load has to supplythe stator copper loss,core loss andfriction and windage loss.The dc resistance of the stator winding is measured and its per phase effective value r1 for AC is calculated from the relation:-r1 = ( ) (dc resistance per phase)For computing the friction and windage loss, the applied voltage to the unloaded induction motor is varied from 20% to about 1.25 of the rated voltage.The input power, current and voltage are recorded so that a graph can be plotted.The speed with reduced voltage, will fall only slightly so that the friction and windage loss remains substantially constant.
32Prot VS V r1 - is the effective per phase stator resistance. From the input-power readings, the corresponding stator ohmic loss is subtracted to obtain the core loss and friction and windage loss ( rotational loss); i.e.Where, Pnl - is the per phase power input,Inl - is the per phase stator current andr1 - is the effective per phase stator resistance.The plot of the rotational loss Prot with variable stator voltage is shown in fig below.Core loss at rated voltageProt VS VFric. & Windage lossFig. 4.16
33Friction & windage loss (Pf&w) The intercept of the extraplotted Prot curve with the ordinate gives the friction and windage loss, because the core loss is zero for zero applied voltage.In order to get a motor accurate value of mechanical loss (friction and windage loss), rotational loss Prot should be plotted against (Voltage)2.This plot of Prot with (voltage)2 is almost linear and, therefore, the extrapolation is easier.ProtProt Vs V2Core loss atrated voltage (Pc)Friction & windage loss (Pf&w)V2Vrat.Fig. 4.17