Presentation on theme: "4. Equivalent circuit of IM. Developing an equivalent circuit model is useful to study and predict the performance characteristics of the induction machine."— Presentation transcript:
4. Equivalent circuit of IM. Developing an equivalent circuit model is useful to study and predict the performance characteristics of the induction machine with reasonable accuracy. In this section a steady-state per-phase equivalent circuit will be derived. For convenience, consider a three-phase wound-rotor induction machine as shown in fig 3.1. Fig phase induction machines equivalent circuit model
If currents flow in both stator and rotor windings, resultant rotating magnetic fields will be produced in the air gap. This resultant air gap field will induce voltages in both stator windings (at supply frequency f 1 ) and rotor windings (at slip frequency f 2 ). It appears that the equivalent circuit may assume a form identical to that of a transformer; therefore it can be developed from the knowledge of a transformer equivalent circuit and from knowledge of variation of rotor 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.
The stator circuit model As explained earlier, when a voltage V 1 is applied to the stator terminals, the rotating flux generates counter emf in the 3-phase of stator windings. The applied voltage V 1 has to overcome counter emf = - E 1, V 1 ` and stator leakage impedance drop,I 1 (r 1 + jx 1 ). Thus, Fig The stator circuit model. V1 = per-phase terminal voltage R1 = per-phase stator winding resistance X1 = per-phase stator leakage reactance E1 = per-phase induced voltage in the stator winding Xm = per-phase stator magnetizing reactance Rc = per-phase stator core loss resistance
The stator current consists of two components, the load component, I 2 ` (-I 2 ) and the exciting component I e. I 2 ` counteracts the rotor mmf, I 2 N 2. The exciting component I e, 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 I 2 R losses I 0 is usually 30% - 50% of rated current of IM. Thus, the exciting current can be resolved into two components. - magnetising component, Im lagging V 1 ` by 90 0 and - core loss component I c which is in phase with V 1 `. In the equivalent circuit, I c and I m are accounted for by a shunt branch consisting of:- - core loss resistance R c = V 1 `/I c - magnetising reactance X m = V 1 `/I m
Since there is air-gap between stator and rotor of induction motor, X m (reactance on account of effect of magnetizing current) will be much less than transformer case. Thus, I m will be much greater than in IM. F, AT Ф, wb Transformer Induction motor Note:- The graph indicates that, Reluctance of IM is much greater than reluctance of transformers. X m = ω s L s = 2πf 1 L s L s =(N s ) 2 /R L Fig. 4.3 Flux mmf relation in transformer and IM.
Rotor circuit model Regarding the actual physical facts existing in the rotor circuit, the per phase rotor current is: Where, SE 2 – slip frequency emf I 2 – slip frequency current If the right hand side is divided by slip,we get:- ( for further analysis ) Fig. 4.4a Rotor equiv. circuit at slip s. Fig. 4.4b I2I2 E 2 – per phase equi. Circuit at stator frequency.
To determine the complete equivalent circuit of the rotor, the line frequency emf E 2, the line frequency current I 2 and the speed dependant resistance r 2 /s must be referred to the stator primary by the effective turn's ratio as in the transformer. Where, N 1 ` = N 1 K w1 N 2 ` = N 2 K w2
Fig.4.5a Rotor equivalent circuit referred to the stator fig.4.5b Exact equivalent circuit of IM Where,
The equivalent circuit of IM is almost similar to the transformer equivalent circuit. i.e. - when S = 1, the exact equivalent circuit becomes the equivalent circuit of a short circuited transformer. - In case S = 0, (at synchronous speed), the circuit becomes the equivalent circuit of an open circuited transformer.
Analysis 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.
Approximate 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
This approximation of the equivalent circuit will considerably simplify computation, because the excitation current (I e ) and the load component (I' 2 ) of the machine can be directly computed from the terminal voltage V 1 by dividing it by the corresponding impedance. Note that if the induction machine is connected to a supply of fixed voltage and frequency, the stator core loss is fixed. Core loss = (hyst. + eddy current) losses hyst. Loss voltage applied eddy loss ( f ) 2 At no load, the machine will operate close to synchronous speed. Therefore, the rotor frequency f 2 is very small and hence rotor core loss is very small. At a lower speed f 2 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 are maximum and as speed falls these losses decreases.
Therefore, 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, R c can be removed from the equivalent circuit, as shown in the fig. below. Fig. 4.7
IEEE RECOMMENDED EQUIVALENT CIRCUIT In the induction machine, because of its air gap, the exciting current I e is high of the order of 30 to 50 percent of the full-load current. The leakage Reactance X 1 is also high. The IEEE recommends that, in such situation, the magnetizing reactance Xm can not be moved to the machine terminals (as is done in transformers.), but be retained at its appropriate place, as shown in the fig.3.8 below. The resistance RC is however, omitted, and the core loss is lumped with the windage and friction losses. This equivalent circuit is to be preferred for situation in which the induced voltage E1 differs appreciably from the terminal voltage V1. Fig. 4.8 a. Equivalent circuit of IM without core loss. ZfZf a b
Steady state performance parameters of IM, such as current, speed, torque, losses, etc. can be computed from the circuit of fig In this fig., Z f is the per phase impedance offered by the rotating air-gap field. Note that Z f includes ZfZf a Fig 4.8b
Impedance Z 1 as seen by stator voltage V 1 is:- And The total power lost in The power lost in therefore,
Determining 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.
The No-load test The induction motor is made to run at no load at rated voltage and frequency. Per phase values of applied stator voltage V nl, input current I nl and input power P nl are recorded ( by two wattmeter method). Fig. 4.9 Circuit diagram for no-load and blocked-rotor Test Fig IEEE RECOMMENDED EQUIVALENT CIRCUIT EQUIVALENT CIRCUIT
Since slip S nl at no-load is very small, r 2 /s at no-load is very large as compared to X m. Thus, the total impedance of parallel branches consisting of jX m and (r 2 /s + jX 2 ) is almost equal to jX m. Then, the no-load reactance X nl seen from the stator terminals is:- jX nl = jX 1 + jX m = X 1 – stator self reactance. Fig Equivalent circuit under no-load test.
From no-load test readings we can get;,, 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. If X 1 is known, X m can be determined.
Thus, The no-load test measures the rotational losses and gives information about its magnetisation current. i.e., no-load test gives X nl and P rot Note that, the only load during no-load test is friction and windage loss. The mechanical power (P m = P g - P rcul ) is consumed by mechanical losses.
PMPM Fig Power flow diagram of IM. Stator I 2 R loss Stator core loss Rotor I 2 R loss Rotor core loss Friction and windage loss PgPg Rotor input power Shaft power Mech. power developed Input power (3IVcosθ)
DC test for stator resistance The rotor resistance r 2 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 r 2, it is necessary to find r 1 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 is measured. If the supply dc voltage is connected between the two phases, ( the 3 rd phase being open) then, the dc current flows through the two phase windings and the required per phase stator resistance is:-
DC w A V + _ R S T Fig 4.13 Circuit for Dc test to determine stator resistance
Blocked rotor test During this test; - An Ac voltage is applied to the stator; and the current is 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 r 2 /s is just equal to r 2 which is quite a small value. Fig IEEE RECOMMENDED EQUIVALENT CIRCUIT
Compared with X m, r 2 & X 2 are small, so that the whole input current flows through them. The equivalent circuit under this condition looks a series combination of r 1, X 1, X 2, r 2. I br Fig.4.15 equivalent circuit during blocked rptor test
Impedance of the IM during block rotor test is:- Where, X br is the reactance of stator & rotor In practice, it is difficult to get separately stator and rotor reactance, and are usually taken from experimental data from tables.
However, for wound rotor machines x 1 is assumed to be equal to x 2. i.e. x 1 = x 2 = ½X br For squirrel cage induction machines, total leakage reactance X br (=x 1 + x 2 ) can be distributed between stator and rotor as per the following table: Empirical distribution of leakage reactance X br Class of motorFraction of X br X1X1 X2X2 1.Class A (normal T st, normal I st and low slip) 2.Class B (normal T st, low I st and low slip) 3.Class C (high T st, low I st and low slip) 4.Class D (high T st, low I st and high slip) For design classes, please refer Stephen J. Chapman, 2 nd ed.
Class of motorFraction of X br X1X1 X2X2 1.Class A (normal T st, normal I st and low slip) 2.Class B (normal T st, low I st and low slip) 3.Class C (high T st, low I st and low slip) 4.Class D (high T st, low I st and high slip)
Thus, the block rotor test helps to determine equivalent circuit parameters r 2, X br (X 1 +X 2 ). If X 1 & X 2 are taken from tables, then X m can be determined. i. e.; X nl = X 1 + X m In 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.
Separation of friction and windage loss from the no-load test result The power input to the induction motor at no-load has to supply the stator copper loss, core loss and friction and windage loss. The dc resistance of the stator winding is measured and its per phase effective value r 1 for AC is calculated from the relation:- r 1 = ( ) (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.
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, P nl - is the per phase power input, I nl - is the per phase stator current and r 1 - is the effective per phase stator resistance. The plot of the rotational loss P rot with variable stator voltage is shown in fig below. Fig Core loss at rated voltage Fric. & Windage loss P rot VS V
The intercept of the extraplotted P rot 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 P rot should be plotted against (Voltage) 2. This plot of Prot with (voltage) 2 is almost linear and, therefore, the extrapolation is easier. P rot V2V2 V rat. Friction & windage loss (P f&w ) Core loss at rated voltage (P c ) P rot Vs V 2 Fig. 4.17