Three phase induction motor basics

three phase induction motor construction and working ppt three phase induction motor dynamic mathematical model
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Published Date:14-07-2017
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Introduction to 3 phase induction motors Definition: Induction motor is an a.c. motor in which currents in the stator winding (which is connected to the supply ) set up a flux which causes currents to be induced in the rotor winding ; these currents interact with the flux to produce rotation . Also called asynchronous motor. The rotor receives electrical power in exactly the same way as the secondary of a two winding transformer receiving its power from primary. That is why an induction motor can be called as a rotating transformer i.e., in which primary winding is stationary but the secondary is free to rotate. Types of induction motors: Depending on rotor there are two types of induction motors :- 1- Squirrel cage induction motor : Rotors is very simple and consist of bars of aluminum (or copper) with shorting rings at the ends. 2- wound rotor induction motor : Three phase windings (star connected) with terminals brought out to slip rings for external connections. The first type is more common used compared to second one due to: 1- Robust : No brushes . No contacts on rotor shaft. 2- Easy to manufacture. 3- Almost maintenance-free , except for bearing and other mechanical parts. 4- Since the rotor has very low resistance, the copper loss is low and efficiency is high . Construction: There are two main types of components which are used in induction motor manufacturing as follows: 1- Active components : which are classified into two categories:- a- Magnetic materials (0.5 mm electrical steel) b- Electrical materials ( copper wires ,insulations ,bars , end rings ,slip rings ,brushes , and lead wires) 1 2- Constructional components: like frame , end shields , shaft , bearings , and fan . These components are shown in Figures 1 & 2 . Figure- 1 parts of squirrel cage induction motor Figure- 2 Axial view of squirrel cage induction motor 2 Stator construction : The stator is made up of several thin laminations (0.5 mm )of electrical steel (silicon steel) , they are punched and clamped together to form a hollow cylinder ( stator core ) with slots , as shown in Figure- 3. Coils of insulated wires are inserted into these slots . Each grouping of coils , together with the core it surrounds , forms an electromagnet (a pair of poles).The number of poles of an induction motor depends on the internal connection of the stator windings . Rotor construction: The squirrel cage rotor is made up of several thin electrical steel lamination (0.5mm) with evenly spaced bars , which are made up of aluminum or copper , along the periphery .In the most popular type of rotor (squirrel cage rotor) , these bars are connected at ends mechanically and electrically by the use of end rings as shown in Fig.4. Almost 90 % of induction motors have squirrel cage rotors . This is because the squirrel cage rotors has a simple and rugged construction .The rotor slots are not exactly parallel to the shaft. Instead , they are given a skew for two main reasons :The first reason is to make the motor run quietly by reducing magnetic hum and to decrease slot harmonics. The second reason is to help reduce the locking tendency of the rotor (the rotor teeth tend to remain locked under the stator teeth due to direct magnetic attraction between the two). The rotor is mounted on the shaft using bearings on each end ; one end of the shaft is normally kept longer than the other for driving the load . Between the stator and the rotor, there exist an air gap , through which due to induction , the energy is transferred from the stator to the rotor. 3 The wound rotor has a set of windings on the rotor slots which are not short circuited , but are terminated to a set of slip rings . These are helpful in adding external resistors and contactors , as shown in Figure 5. 4 Typical name plate of induction motor A typical name plate of induction motor is shown in Figure 6, and Table 1 . Motor standards NEMA : National Electrical Manufactures Association IEC : International Electrotechnical Commission 5 Motor insulation class Insulations have been standardized and graded by their resistance to thermal aging and failure. Four insulation classes are in common use , they have been designated by the letters A , B , F, and H . The temperature capabilities of these classes are separated from each other by 25 °C increments. The temperature capabilities of each insulation class is defined as being the maximum temperature at which the insulation can be operated to yield an average life of 20,000 hours , as in Table below. Insulation Class Temperature Rating A 105° C B 130° C F 155° C H 180° C Motor degree of protection I P : International Protection , I P Protection against ingress Protection against ingress of bodies of water 0 Non protected 0 Non protected 1 Protected against ingress of 1 Protected against ingress of foreign solid bodies of 50 dripping water . mm or greater. 2 Protected against ingress of 2 Protection against ingress of foreign solid bodies of 12 dripping water at maximum mm or greater. angle of 150 degrees from the vertical. 3 Protected against ingress of 3 Protection against water foreign solid bodies of 2.5 falling like rain. mm or greater. 4 Protected against ingress of 4 Protection against splashing foreign solid bodies of 1 mm water. or greater. 5 Partially protected against 5 Protection against water jets ingress of dust . . 6 Totally protected against 6 Protection against special ingress of dust. conditions on ship's board. 7 Protection against immersion in water . 8 Protection against prolonged immersion in water . 6Chapter (8) Three Phase Induction Motors Introduction The three-phase induction motors are the most widely used electric motors in industry. They run at essentially constant speed from no-load to full-load. However, the speed is frequency dependent and consequently these motors are not easily adapted to speed control. We usually prefer d.c. motors when large speed variations are required. Nevertheless, the 3-phase induction motors are simple, rugged, low-priced, easy to maintain and can be manufactured with characteristics to suit most industrial requirements. In this chapter, we shall focus our attention on the general principles of 3-phase induction motors. 8.1 Three-Phase Induction Motor Like any electric motor, a 3-phase induction motor has a stator and a rotor. The stator carries a 3-phase winding (called stator winding) while the rotor carries a short-circuited winding (called rotor winding). Only the stator winding is fed from 3-phase supply. The rotor winding derives its voltage and power from the externally energized stator winding through electromagnetic induction and hence the name. The induction motor may be considered to be a transformer with a rotating secondary and it can, therefore, be described as a “transformer- type” a.c. machine in which electrical energy is converted into mechanical energy. Advantages (i) It has simple and rugged construction. (ii) It is relatively cheap. (iii) It requires little maintenance. (iv) It has high efficiency and reasonably good power factor. (v) It has self starting torque. Disadvantages (i) It is essentially a constant speed motor and its speed cannot be changed easily. (ii) Its starting torque is inferior to d.c. shunt motor. 1818.2 Construction A 3-phase induction motor has two main parts (i) stator and (ii) rotor. The rotor is separated from the stator by a small air-gap which ranges from 0.4 mm to 4 mm, depending on the power of the motor. 1. Stator It consists of a steel frame which encloses a hollow, cylindrical core made up of thin laminations of silicon steel to reduce hysteresis and eddy current losses. A number of evenly spaced slots are provided on the inner periphery of the laminations See Fig. (8.1). The insulated connected to form a Fig.(8.1) balanced 3-phase star or delta connected circuit. The 3-phase stator winding is wound for a definite number of poles as per requirement of speed. Greater the number of poles, lesser is the speed of the motor and vice-versa. When 3-phase supply is given to the stator winding, a rotating magnetic field (See Sec. 8.3) of constant magnitude is produced. This rotating field induces currents in the rotor by electromagnetic induction. 2. Rotor The rotor, mounted on a shaft, is a hollow laminated core having slots on its outer periphery. The winding placed in these slots (called rotor winding) may be one of the following two types: (i) Squirrel cage type (ii) Wound type (i) Squirrel cage rotor. It consists of a laminated cylindrical core having parallel slots on its outer periphery. One copper or aluminum bar is placed in each slot. All these bars are joined at each end by metal rings called end rings See Fig. (8.2). This forms a permanently short-circuited winding which is indestructible. The entire construction (bars and end rings) resembles a squirrel cage and hence the name. The rotor is not connected electrically to the supply but has current induced in it by transformer action from the stator. Those induction motors which employ squirrel cage rotor are called squirrel cage induction motors. Most of 3-phase induction motors use squirrel cage rotor as it has a remarkably simple and robust construction enabling it to operate in the most adverse circumstances. However, it suffers from the disadvantage of a low starting torque. It is because the rotor bars are permanently short-circuited and it is not possible to add any external resistance to the rotor circuit to have a large starting torque. 182Fig.(8.2) Fig.(8.3) (ii) Wound rotor. It consists of a laminated cylindrical core and carries a 3- phase winding, similar to the one on the stator See Fig. (8.3). The rotor winding is uniformly distributed in the slots and is usually star-connected. The open ends of the rotor winding are brought out and joined to three insulated slip rings mounted on the rotor shaft with one brush resting on each slip ring. The three brushes are connected to a 3-phase star-connected rheostat as shown in Fig. (8.4). At starting, the external resistances are included in the rotor circuit to give a large starting torque. These resistances are gradually reduced to zero as the motor runs up to speed. Fig.(8.4) The external resistances are used during starting period only. When the motor attains normal speed, the three brushes are short-circuited so that the wound rotor runs like a squirrel cage rotor. 8.3 Rotating Magnetic Field Due to 3-Phase Currents When a 3-phase winding is energized from a 3-phase supply, a rotating magnetic field is produced. This field is such that its poles do no remain in a fixed position on the stator but go on shifting their positions around the stator. For this reason, it is called a rotating Held. It can be shown that magnitude of this rotating field is constant and is equal to 1.5 ϕ where ϕ is the maximum m m flux due to any phase. 183To see how rotating field is produced, consider a 2-pole, 3i-phase winding as shown in Fig. (8.6 (i)). The three phases X, Y and Z are energized from a 3-phase source and currents in these phases are indicated as I , I and I x y z See Fig. (8.6 (ii)). Referring to Fig. (8.6 (ii)), the fluxes produced by these currents are given by: Fig.(8.5) ϕ = ϕ sin ωt x m ϕ = ϕ sin (ωt -120°) y m ϕ = ϕ sin (ωt - 240°) z m Here ϕ is the maximum flux due to any phase. Fig. (8.5) shows the phasor m diagram of the three fluxes. We shall now prove that this 3-phase supply produces a rotating field of constant magnitude equal to 1.5 ϕm. Fig.(8.6) 184(i) At instant 1 See Fig. (8.6 (ii)) and Fig. (8.6 (iii)), the current in phase X is zero and currents in phases Y and Z are equal and opposite. The currents are flowing outward in the top conductors and inward in the bottom conductors. This establishes a resultant flux towards right. The magnitude of the resultant flux is constant and is equal to 1.5 ϕ as m proved under: Fig.(8.7) At instant 1, ωt = 0°. Therefore, the three fluxes are given by; 3 ϕ = 0; ϕ = ϕ sin(-120°) = - ϕ ; x y m m 2 3 ϕ = ϕ sin(- 240°) = ϕ z m m 2 The phasor sum of - ϕ and ϕ is the resultant flux ϕ See Fig. (8.7). It is y z r clear that: 3 60° 3 3 Resultant flux, ϕ = 2× ϕ cos = 2× ϕ × =1.5 ϕ r m m m 2 2 2 2 (ii) At instant 2, the current is maximum (negative) in ϕ phase Y and 0.5 maximum y (positive) in phases X and Y. The magnitude of resultant flux is 1.5 ϕ as m proved under: At instant 2, ωt = 30°. Therefore, the three fluxes are given by; Fig.(8.8) ϕ m ϕ = ϕ sin 30° = x m 2 ϕ = ϕ sin (-90°) = -ϕ y m m ϕ m ϕ = ϕ sin (-210°) = z m 2 The phasor sum of ϕ , - ϕ and ϕ is the resultant flux ϕ x y z r ϕ ϕ 120° m m Phasor sum of ϕ and ϕ , ϕ' = 2 × cos = x z r 2 2 2 ϕ m Phasor sum of ϕ' and - ϕ , ϕ = + ϕ =1.5 ϕ r y r m m 2 Note that resultant flux is displaced 30° clockwise from position 1. 185(iii) At instant 3, current in phase Z is zero and the currents in phases X and Y are equal and opposite (currents in phases X and Y arc 0.866 × max. value). The magnitude of resultant flux is 1.5 ϕ as proved under: m At instant 3, ωt = 60°. Therefore, the three fluxes are given by; Fig.(8.9) 3 ϕ = ϕ sin 60° = ϕ ; x m m 2 3 ϕ = ϕ sin(- 60°) = - ϕ ; y m m 2 ϕ = ϕ sin(-180°)= 0 z m The resultant flux ϕ is the phasor sum of ϕ and - ϕ(Θ ϕ = 0). r x y z 3 60° ϕ = 2 × ϕ cos =1.5 ϕ r m m 2 2 Note that resultant flux is displaced 60° clockwise from position 1. (iv) At instant 4, the current in phase X is maximum (positive) and the currents in phases V and Z are equal and negative (currents in phases V and Z are 0.5 × max. value). This establishes a resultant flux downward as shown under: Fig.(7.10) At instant 4, ωt = 90°. Therefore, the three fluxes are given by; ϕ = ϕ sin90° = ϕ x m m ϕ m ϕ = ϕ sin (-30°) = - y m 2 ϕ m ϕ = ϕ sin (-150°) = - z m 2 The phasor sum of ϕ , - ϕ and - ϕ is the resultant flux ϕ x y z r ϕ ϕ 120° m m Phasor sum of - ϕ and - ϕ , ϕ' = 2 × cos = z y r 2 2 2 ϕ m Phasor sum of ϕ' and ϕ , ϕ = + ϕ =1.5 ϕ r x r m m 2 Note that the resultant flux is downward i.e., it is displaced 90° clockwise from position 1. 186It follows from the above discussion that a 3-phase supply produces a rotating field of constant value (= 1.5 ϕ , where ϕ is the maximum flux due to any m m phase). Speed of rotating magnetic field The speed at which the rotating magnetic field revolves is called the synchronous speed (N ). Referring to Fig. (8.6 (ii)), the time instant 4 represents s the completion of one-quarter cycle of alternating current I from the time x instant 1. During this one quarter cycle, the field has rotated through 90°. At a time instant represented by 13 or one complete cycle of current I from the x origin, the field has completed one revolution. Therefore, for a 2-pole stator winding, the field makes one revolution in one cycle of current. In a 4-pole stator winding, it can be shown that the rotating field makes one revolution in two cycles of current. In general, fur P poles, the rotating field makes one revolution in P/2 cycles of current. P ∴ Cycles of current = × revolutions of field 2 P or Cycles of current per second = × revolutions of field per second 2 Since revolutions per second is equal to the revolutions per minute (N ) divided s by 60 and the number of cycles per second is the frequency f, N N P P s s ∴ f = × = 2 60 120 120 f or N = s P The speed of the rotating magnetic field is the same as the speed of the alternator that is supplying power to the motor if the two have the same number of poles. Hence the magnetic flux is said to rotate at synchronous speed. Direction of rotating magnetic field The phase sequence of the three-phase voltage applied to the stator winding in Fig. (8.6 (ii)) is X-Y-Z. If this sequence is changed to X-Z-Y, it is observed that direction of rotation of the field is reversed i.e., the field rotates counterclockwise rather than clockwise. However, the number of poles and the speed at which the magnetic field rotates remain unchanged. Thus it is necessary only to change the phase sequence in order to change the direction of rotation of the magnetic field. For a three-phase supply, this can be done by interchanging any two of the three lines. As we shall see, the rotor in a 3-phase induction motor runs in the same direction as the rotating magnetic field. Therefore, the 187direction of rotation of a 3-phase induction motor can be reversed by interchanging any two of the three motor supply lines. 8.4 Alternate Mathematical Analysis for Rotating Magnetic Field We shall now use another useful method to find the magnitude and speed of the resultant flux due to three-phase currents. The three-phase sinusoidal currents produce fluxes ϕ , ϕ and ϕ which vary 1 2 3 sinusoidally. The resultant flux at any instant will be the vector sum of all the three at that instant. The fluxes are represented by three variable Fig.(8.11) magnitude vectors See Fig. (8.11). In Fig. (8.11), the individual flux directions arc fixed but their magnitudes vary sinusoidally as does the current that produces them. To find the magnitude of the resultant flux, resolve each flux into horizontal and vertical components and then find their vector sum. ϕ = ϕ cost ωt - ϕ cos(ωt -120°)cos60° - ϕ cos(ωt - 240°)cos60° h m m m 3 = ϕ cosωt m 2 3 ϕ = 0 - ϕ cos(ωt -120°)cos60° + ϕ cos(ωt - 240°)cos60° = ϕ sin ωt v m m m 2 The resultant flux is given by; 1/ 2 3 3 2 2 2 2 ϕ = ϕ + ϕ = ϕcos ωt + (-sin ωt) = ϕ =1.5ϕ = Constant r h v m m m 2 2 Thus the resultant flux has constant magnitude (= 1.5 ϕ ) and does not change m with time. The angular displacement of ϕ relative to r the OX axis is 3 ϕ sin ωt m ϕ v 2 tan θ = = = tan ωt ϕ 3 h ϕ cosωt m Fig.(8.12) 2 ∴ θ = ωt Thus the resultant magnetic field rotates at constant angular velocity ω( = 2 πf) rad/sec. For a P-pole machine, the rotation speed (ω ) is m 2 ω = ω rad/sec m P 1882πN 2 s or = × 2πf ... N is in r.p.m. s 60 P 120 f ∴ N = s P Thus the resultant flux due to three-phase currents is of constant value (= 1.5 ϕ m where ϕ is the maximum flux in any phase) and this flux rotates around the m stator winding at a synchronous speed of 120 f/P r.p.m. For example, for a 6-pole, 50 Hz, 3-phase induction motor, N, = 120 × 50/6 = 1000 r.p.m. It means that flux rotates around the stator at a speed of 1000 r.p.m. 8.5 Principle of Operation Consider a portion of 3-phase induction motor as shown in Fig. (8.13). The operation of the motor can be explained as under: (i) When 3-phase stator winding is energized from a 3-phase supply, a rotating magnetic field is set up which rotates round the stator at synchronous speed N (= 120 f/P). s (ii) The rotating field passes through the air gap and cuts the rotor conductors, Fig.(1-) which as yet, are stationary. Due to the relative speed between the rotating flux and the stationary rotor, e.m.f.s are induced in the rotor conductors. Since the rotor circuit is short-circuited, currents start flowing in the rotor conductors. (iii) The current-carrying rotor conductors are placed in the magnetic field produced by the stator. Consequently, mechanical force acts on the rotor conductors. The sum of the mechanical forces on all the rotor conductors produces a torque which tends to move the rotor in the same direction as the rotating field. (iv) The fact that rotor is urged to follow the stator field (i.e., rotor moves in the direction of stator field) can be explained by Lenz’s law. According to this law, the direction of rotor currents will be such that they tend to oppose the cause producing them. Now, the cause producing the rotor currents is the relative speed between the rotating field and the stationary rotor conductors. Hence to reduce this relative speed, the rotor starts running in the same direction as that of stator field and tries to catch it. 1898.6 Slip We have seen above that rotor rapidly accelerates in the direction of rotating field. In practice, the rotor can never reach the speed of stator flux. If it did, there would be no relative speed between the stator field and rotor conductors, no induced rotor currents and, therefore, no torque to drive the rotor. The friction and windage would immediately cause the rotor to slow down. Hence, the rotor speed (N) is always less than the suitor field speed (N ). This difference s in speed depends upon load on the motor. The difference between the synchronous speed N of the rotating stator field and s the actual rotor speed N is called slip. It is usually expressed as a percentage of synchronous speed i.e., N - N s % age slip, s = ×100 N s (i) The quantity N - N is sometimes called slip speed. s (ii) When the rotor is stationary (i.e., N = 0), slip, s = 1 or 100 %. (iii) In an induction motor, the change in slip from no-load to full-load is hardly 0.1% to 3% so that it is essentially a constant-speed motor. 8.7 Rotor Current Frequency The frequency of a voltage or current induced due to the relative speed between a vending and a magnetic field is given by the general formula; NP Frequency = 120 where N = Relative speed between magnetic field and the winding P = Number of poles For a rotor speed N, the relative speed between the rotating flux and the rotor is N - N. Consequently, the rotor current frequency f' is given by; s (N - N)P s f '= 120 s N P N - N ⎛ ⎞ s s = Θ s = ⎜ ⎟ 120 N ⎝ s ⎠ N P ⎛ ⎞ s = sf Θ f = ⎜ ⎟ 120 ⎝ ⎠ i.e., Rotor current frequency = Fractional slip x Supply frequency (i) When the rotor is at standstill or stationary (i.e., s = 1), the frequency of rotor current is the same as that of supply frequency (f' = sf = 1× f = f). 190(ii) As the rotor picks up speed, the relative speed between the rotating flux and the rotor decreases. Consequently, the slip s and hence rotor current frequency decreases. Note. The relative speed between the rotating field and stator winding is N - 0 s = N. Therefore, the frequency of induced current or voltage in the stator s winding is f = N P/120—the supply frequency. s 8.8 Effect of Slip on The Rotor Circuit When the rotor is stationary, s = 1. Under these conditions, the per phase rotor e.m.f. E has a frequency equal to that of supply frequency f. At any slip s, the 2 relative speed between stator field and the rotor is decreased. Consequently, the rotor e.m.f. and frequency are reduced proportionally to sE and sf respectively. s At the same time, per phase rotor reactance X , being frequency dependent, is 2 reduced to sX . 2 Consider a 6-pole, 3-phase, 50 Hz induction motor. It has synchronous speed N s = 120 f/P = 120 × 50/6 = 1000 r.p.m. At standsill, the relative speed between stator flux and rotor is 1000 r.p.m. and rotor e.m.f./phase = E (say). If the full- 2 load speed of the motor is 960 r.p.m., then, 1000 - 960 s = = 0.04 1000 (i) The relative speed between stator flux and the rotor is now only 40 r.p.m. Consequently, rotor e.m.f./phase is reduced to: 40 E × = 0.04E or sE 2 2 2 1000 (ii) The frequency is also reduced in the same ratio to: 40 50× = 50 ×0.04 or sf 1000 (iii) The per phase rotor reactance X is likewise reduced to: 2 40 X × =0.04X or sX 2 2 2 1000 Thus at any slip s, Rotor e.m.f./phase = sE 2 Rotor reactance/phase = sX 2 Rotor frequency = sf where E ,X and f are the corresponding values at standstill. 2 2 1918.9 Rotor Current Fig. (8.14) shows the circuit of a 3-phase induction motor at any slip s. The rotor is assumed to be of wound type and star connected. Note that rotor e.m.f./phase and rotor reactance/phase are s E and sX respectively. The rotor 2 2 resistance/phase is R2 and is independent of frequency and, therefore, does not depend upon slip. Likewise, stator winding values R and X do not depend 1 1 upon slip. Fig.(8.14) Since the motor represents a balanced 3-phase load, we need consider one phase only; the conditions in the other two phases being similar. At standstill. Fig. (8.15 (i)) shows one phase of the rotor circuit at standstill. E E 2 2 Rotor current/phase, I = = 2 2 2 Z 2 R + X 2 2 R R 2 2 Rotor p.f., cos ϕ = = 2 2 2 Z 2 R + X 2 2 Fig.(8.15) When running at slip s. Fig. (8.15 (ii)) shows one phase of the rotor circuit when the motor is running at slip s. sE sE 2 2 Rotor current, I' = = 2 Z' 2 2 2 R +(sX) 2 2 192R R 2 2 Rotor p.f., cos ϕ' = = 2 Z' 2 2 2 R +(sX) 2 2 8.10 Rotor Torque The torque T developed by the rotor is directly proportional to: (i) rotor current (ii) rotor e.m.f. (iii) power factor of the rotor circuit ∴ T ∝E I cosϕ 2 2 2 or T = KE I cosϕ 2 2 2 where I = rotor current at standstill 2 E = rotor e.m.f. at standstill 2 cos ϕ = rotor p.f. at standstill 2 Note. The values of rotor e.m.f., rotor current and rotor power factor are taken for the given conditions. 8.11 Starting Torque (T ) s Let E = rotor e.m.f. per phase at standstill 2 X = rotor reactance per phase at standstill 2 R = rotor resistance per phase 2 2 2 Rotor impedance/phase, Z = R + X standstill 2 2 2 E E 2 2 Rotor current/phase, standstill I = = 2 Z 2 2 2 R + X 2 2 R R 2 2 Rotor p.f., cos ϕ = = standstill 2 Z 2 2 2 R + X 2 2 ∴ Starting torque,T =KE I cosϕ s 22 2 E R 2 2 =KE × × 2 2 2 2 2 R +X R +X 2 2 2 2 2 KE R 2 2 = 2 2 R +X 2 2 193Generally, the stator supply voltage V is constant so that flux per pole ϕ set up by the stator is also fixed. This in turn means that e.m.f. E induced in the rotor 2 will be constant. K R K R 1 2 1 2 ∴ T = = s 2 2 2 R + X Z 2 2 2 where K is another constant. 1 It is clear that the magnitude of starting torque would depend upon the relative values of R and X i.e., rotor resistance/phase and standstill rotor 2 2 reactance/phase. It can be shown that K = 3/2 π N . s 2 E R 3 2 2 ∴ T = ⋅ s 2 2 2π N R + X s 2 2 Note that here N is in r.p.s. s 8.12 Condition for Maximum Starting Torque It can be proved that starting torque will be maximum when rotor resistance/phase is equal to standstill rotor reactance/phase. K R 1 2 Now T = (i) s 2 2 R + X 2 2 Differentiating eq. (i) w.r.t. R and equating the result to zero, we get, 2 ⎡ ⎤ dT R (2R ) 1 s 2 2 ⎢ ⎥ = K - = 0 1 2 2 2 dR 2 2 2 ⎢R + X ⎥ 2 2 (R + X ) ⎣ 2 2 ⎦ 2 2 2 or R + X = 2R 2 2 2 or R = X 2 2 Hence starting torque will be maximum when: Rotor resistance/phase = Standstill rotor reactance/phase Under the condition of maximum starting torque, ϕ = 45° and rotor power 2 factor is 0.707 lagging See Fig. (8.16 (ii)). Fig. (8.16 (i)) shows the variation of starting torque with rotor resistance. As the rotor resistance is increased from a relatively low value, the starting torque increases until it becomes maximum when R = X . If the rotor resistance is 2 2 increased beyond this optimum value, the starting torque will decrease. 194

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