CHAPTER35 Learning Objectives ➣ General ➣ Circle Diagram for a Series COMPUTATIONS Circuit ➣ Circle Diagram of the Approximate Equivalent AND CIRCLE Circle ➣ Determination of G0 and B0 ➣ No-load Test ➣ Blocked Rotor Test DIAGRAMS ➣ Construction of the Circle Diagram ➣ Maximum Quantities ➣ Starting of Induction Motors ➣ Direct-Switching or Line Starting of Induction Motors ➣ Squirrel-cage Motors ➣ Starting of Slip-ring Motors ➣ Starter Steps ➣ Crawling ➣ Cogging or Magnetic Locking ➣ Double Sqiurrel-cage Motor ➣ Equivalent circuit ➣ Speed Control of Induction Motor ➣ Three-Phase A.C. Commutator Motors ➣ Schrage Motor ➣ Motor Enclosures ➣ Standard type of Squirrel- cage Motors ➣ Class A Motors ➣ Class B Motors This chapter explains you how to derive ➣ Ç Class C Motors performance characteristics of induction ➣ Class D Motors motors using circular diagrams ➣ Class E Motors ➣ Class F Motors ➣ Questions and Answer on Induction Motors 1314 Electrical Technology 35.1. General In this chapter, it will be shown that the performance characteristics of an induction motor are derivable from a circular locus. The data necessary to draw the circle diagram may be found from no- load and blocked-rotor tests, corresponding to the open-circuit and short-circuit tests of a transformer. The stator and rotor Cu losses can be separated by drawing a torque line. The parameters of the motor, in the equivalent circuit, can be found from the above tests, as shown below. 35.2. Circle Diagram for a Series Circuit It will be shown that the end of the current vector for a series circuit with constant reactance and voltage, but with a variable resistance is a circle. With reference to Fig. 35.1, it is clear that VV I = = I X R Z ()RX22+ VXV×=φsin = ZX XX()RX22+ V X ∵ sin φ = – Fig. 35.2 f ()RX22+ R ∴ I =(V/X) sin φ Fig. 35.1 Fig. 35.2 It is the equation of a circle in polar co- ordinates, with diameter equal to V/X. Such a circle is drawn in Fig. 35.3, using the magnitude of the current and power factor angle φ as polar co-ordinates of the point A. In other words, as resistance R φ Y is varied (which means, in fact, is changed), the Y end of the current vector lies on a circle with diameter equal to V/X. For a lagging current, it is usual to orientate the circle of Fig. 35.3 (a) such that its diameter is horizontal and the voltage vector V C A X takes a vertical position, as shown in Fig. 35.3 (b). f I I C f O There is no difference between the two so far as X O X V the magnitude and phase relationships are X concerned. 35.3. Circle Diagram for the Approxi- ()a ()b Fig. 35.3 mate Equivalent Circuit The approximate equivalent diagram is redrawn in Fig. 35.4. It is clear that the circuit to the right of points ab is similar to a series circuit, having a constant voltage V 1 and reactance X 01 but variable resistance (corresponding to different values of slip s). ′ Hence, the end of current vector for I2 will lie on a circle with a diameter of V/X 01. In Fig. 35.5, I ′ is the rotor current referred to stator, I 2 0 XXX¢ is no-load current (or exciting current) and I 01=+ 1 2 RRR¢ aa1 01=+ 1 2 I is the total stator current and is the 1 I¢ ′ I 2 vector sum of the first two. When I2 is 0 φ lagging and 2 = 90º, then the position of ′ V G B vector for I2 will be along OC i.e. at right r 0 0 angles to the voltage vector OE. For any RR¢¢= 1 _ φ L2(S 1 ( other value of 2, point A will move along b the circle shown dotted. The exciting current I0 is drawn lagging V by an angle φ Fig. 35.4 0. If conductance G0 and susceptance Computations and Circle Diagrams 1315 φ B0 of the exciting circuit are assumed constant, then I0 and 0 are also constant. The end of current vector for I1 is also seen to lie on another circle which is displaced from the dotted circle by an amount I0. Its diameter is still V/X 01 and is parallel to the horizontal axis OC. Hence, we find that if an induction motor is tested at various loads, the locus of the end of the vector for the current (drawn by it) is a circle. E 35.4. Determination of G0 and B0 If the total leakage reactance X 01 of the motor, exciting conductance G and exciting 0 V susceptance B are found, then the position 0 B of the circle O′BC′ is determined uniquely. A ¢ One method of finding G and B consists in I 2 0 0 f running the motor synchronously so that slip 2 f I1 ¢ 1 O ¢ s = 0. In practice, it is impossible for an f C 0 V/X induction motor to run at synchronous speed, I0 01 due to the inevitable presence of friction and O V/X C windage losses. However, the induction 01 motor may be run at synchronous speed by Fig. 35.5 I X R another machine which supplies the friction and windage losses. In that case, the circuit to the right of points ab ∞ behaves like an open circuit, because with s = 0, RL = (Fig. 35.6). Hence, the current drawn by the motor is I ZX 0 V only. Let V = applied voltage/phase; I0 = motor current / f phase W = wattmeter reading i.e. input in watt ; Y 0 = R exciting admittance of the motor. Then, for a 3-phase Fig. 35.6 induction motor 2 W W = 3G0V or G0 = Also, I0 = VY0 or Y 0 = I0/V 3V 2 22−= 22 − B0 = ()[(/)]YG00 IVG 0 0 Hence, G0 and B0 can be found. 35.5. No-load Test In practice, it is neither necessary nor feasible to run the induction motor synchronously for getting G0 and B0. Instead, the motor is run without any external mechanical load on it. The speed of the rotor would not be synchronous, but very much near to it ; so that, for all practical purposes, the speed may be assumed synchronous. The no load test is carried out with different values of applied voltage, below and above the value of normal voltage. The power input is measured by two wattmeters, Fig. 35.7 Fig. 35.8 1316 Electrical Technology I0 by an ammeter and V by a voltmeter, which are included in the circuit of Fig. 35.7. As motor is running on light load, the p.f. would be low i.e. less than 0.5, hence total power input will be the difference of the two wattmeter readings W 1 and W 2. The readings of the total power input W 0 , I0 and voltage V are plotted as in Fig. 35.8. If we extend the curve for W 0, it cuts the vertical axis at point A. OA represents losses due to friction and windage. If we subtract loss corresponding to OA from W 0, then we get the no-load electrical and magnetic losses in the machine, because the no-load input W 0 to the motor consists of 2 2 (i) small stator Cu loss 3 I0 R1 (ii) stator core loss W CL = 3G0 V (iii) loss due to friction and windage. The losses (ii) and (iii) are collectively known as fixed losses, because they are independent of load. OB represents normal voltage. Hence, losses at normal voltage can be found by drawing a vertical line from B. BD = loss due to friction and windage DE = stator Cu loss EF = core loss Hence, knowing the core loss W CL, G0 and B0 can be found, as discussed in Art. 35.4. φ φ Additionally, 0 can also be found from the relation W 0 = 3 V L I0 cos 0 W ∴ φ 0 cos 0 = where V L = line voltage and W 0 is no-load stator input. 3 VIL 0 Example 35.1. In a no-load test, an induction motor took 10 A and 450 watts with a line voltage of 110 V. If stator resistance/phase is 0.05 Ω and friction and windage losses amount to 135 watts, calculate the exciting conductance and susceptance/phase. 2 × 2 × Solution. stator Cu loss = 3 I0 R1 = 3 10 0.05 = 15 W ∴ stator core loss = 450 − 135 − 15 = 300 W 2 Voltage/phase V = 110/ 3 V ; Core loss = 3 G0 V × 2 300 300 = 3 G0 (110/ 3 ) ; G0 = 3× (110/ 3)2 = 0.025 siemens/phase × Y0 = I0 / V = (10 3 )/110 = 0.158 siemens/ phase 22−= 2 − 2 B0 = (YG00 ) (0.158 0.025 ) = 0.156 siemens/phase. 35.6. Blocked Rotor Test It is also known as locked-rotor or short-circuit test. This test is used to find– 1. short-circuit current with normal voltage applied to sta- tor 2. power factor on short-circuit Both the values are used in the construction of circle dia- gram This vertical test stand is capable of absorbing up to 10,000 N-m of 3. total leakage reactance X 01 of the motor as referred to pri- torque at continuous load rating mary (i.e. stator) (max 150.0 hp at 1800 rpm). It helps 4. total resistance of the motor R as referred to primary. to develop speed torque curves and 01 performs locked rotor testing In this test, the rotor is locked (or allowed very slow rotation) Computations and Circle Diagrams 1317 and the rotor windings are short-circuited at slip-rings, if the motor has a wound rotor.