Skip to main content

Double Revolving Field Theroy

According to this theory, any alternating quantity can be resolved into two rotating components which rotate in opposite directions and each having magnitude as half of the maximum magnitude of the alternating quantity.
       In case of single phase induction motors, the stator winding produces an alternating magnetic field having maximum magnitude of Φ1m.
       According to double revolving field theory, consider the two components of the stator flux, each having magnitude half of maximum magnitude of stator flux i.e. (Φ1m/2). Both these components are rotating in opposite directions at the synchronous speed Ns which is dependent on frequency and stator poles.
       Let Φ is forward component rotating in anticlockwise direction while Φ is the backward component rotating in clockwise direction. The resultant of these two components at any instant gives the instantaneous value of the stator flux at the instant. So resultant of these two is the original stator flux.

Fig. 1  Stator flux and its two components
       The Fig. 1 shows the stator flux and its two components Φand Φb. At start both the components are shown opposite to each other in the Fig.1(a). Thus the resultant ΦR = 0. This is nothing but the instantaneous value of the stator flux at start. After 90o , as shown in the Fig. 1(b), the two components are rotated in such a way that both are pointing in the same direction. Hence the resultant ΦR is the algebraic sum of the magnitudes of the two components. So ΦR = (Φ1m/2) + (Φ1m/2) =Φ1m. This is nothing but the instantaneous value of the stator flux at θ = 90o as shown in the Fig 1(c). Thus continuous rotation of the two components gives the original alternating stator flux.
       Both the components are rotating and hence get cut by the motor conductors. Due to cutting of flux, e.m.f. gets induced in rotor which circulates rotor current. The rotor current produces rotor flux. This flux interacts with forward component Φto produce a torque in one particular direction say anticlockwise direction. While rotor flux interacts with backward component Φb to produce a torque in the clockwise direction. So if anticlockwise torque is positive then clockwise torque is negative.
       At start these two torque are equal in magnitude but opposite in direction. Each torque tries to rotate the rotor in its own direction. Thus net torque experienced by the rotor is zero at start. And hence the single phase induction motors are not self starting.

Torque speed characteristics
       The two oppositely directed torques and the resultant torque can be shown effectively with the help of  torque-speed characteristics. It is shown in the Fig.2.
Fig. 2  Torque-speed characteristic

       It can be seen that at start N = 0 and at that point resultant torque is zero. So single phase motors are not self starting.
       However if the rotor is given an initial rotation in any direction, the resultant average torque increase in the direction in which rotor initially rotated. And motor starts rotating in that direction. But in practice it is not possible to give initial torque to rotor externally hence some modifications are done in the construction of single phase induction motors to make them self starting.
       Another theory which can also be used to explain why single phase induction motors is not self starting is cross-field theory.

Comments

Popular posts from this blog

Transformer multiple choice questions part 1

Hello Engineer's Q.[1] A transformer transforms (a) frequency (b) voltage (c) current (d) voltage and current Ans : D Q.[2] Which of the following is not a basic element of a transformer ? (a) core (b) primary winding (c) secondary winding (d) mutual flux. Ans : D Q.[3] In an ideal transformer, (a) windings have no resistance (b) core has no losses (c) core has infinite permeability (d) all of the above. Ans : D Q.[4] The main purpose of using core in a transformer is to (a) decrease iron losses (b) prevent eddy current loss (c) eliminate magnetic hysteresis (d) decrease reluctance of the common magnetic circuit. Ans :D Q.[5] Transformer cores are laminated in order to (a) simplify its construction (b) minimize eddy current loss (c) reduce cost (d) reduce hysteresis loss. Ans : B Q.[6] A transformer having 1000 primary turns is connected to a 250-V a.c. supply. For a secondary voltage of 400 V, the number of secondary turns should be (a) 1600 (b) 250 (c) 400 (d) 1250 A

Condition for Maximum Power Developed In Synchronous Motor

The value of δ for which the mechanical power developed is maximum can be obtained as, Note : Thus when R a is negligible, θ = 90 o for maximum power developed. The corresponding torque is called pull out torque. 1.1 The Value of Maximum Power Developed        The value of maximum power developed can be obtained by substituting θ = δ in the equation of P m .        When R a is negligible,     θ = 90 o  and cos (θ) = 0 hence, . . .               R a = Z s cosθ   and X s = Z s sinθ        Substituting   cosθ = R a /Z s in equation (6b) we get,         Solving the above quadratic in E b we get,        As E b is completely dependent on excitation, the equation (8) gives the excitation limits for any load for a synchronous motor. If the excitation exceeds this limit, the motor falls out of step. 1.2 Condition for Excitation When Motor Develops ( P m ) R max        Let us find excitation condition for maximum power developed. The excitation

Effect of Slip on Rotor Parameters : Part2

Effect of Slip on Rotor Parameters 2. Effect of Slip on Magnitude of Rotor Induced E.M.F        We have seen that when rotor is standstill, s  = 1, relative speed is maximum and maximum e.m.f. gets induced in the rotor. Let this e.m.f. be,                 E 2 = Rotor induced e.m.f. per phase on standstill condition         As rotor gains speed, the relative speed between rotor and rotating magnetic field decreases and hence induced e.m.f. in rotor also decreases as it is proportional to the relative speed N s - N. Let this e.m.f. be,                E 2r = Rotor induced e.m.f. per phase in running condition  Now        E 2r α N s while E 2r α N s - N        Dividing the two proportionality equations,               E 2r /E 2 = ( N s - N)/N s    but (N s - N)/N = slip s               E 2r /E 2 = s               E 2r = s E 2        The magnitude of the induced e.m.f in the rotor also reduces by slip times the