Chapter Ten Braking of Three Phase IM

1. Chapter Ten Braking of Three Phase IM During acceleration all electrical motors draw extra energy than the required amount for meeting the load condition. This energy is stored in the moving parts of the machine. This energy which causes the machine to run even after switching off. Some applications may require the motor to stop at a time quicker than in natural. A most common method to achieve this is to apply mechanical braking by means of friction offered by brake-shoes in which stored energy is dissipated in the form of heat. But this method gives rise to wear and requires periodical replacement of brake-shoes also.

2. 1- The Plugging: The braking torque is provided by interchanging two of three supply terminals, so that the direction of the rotating field is reversed. The electromagnetic torque thus developed is opposite to motoring torque, which bring the motor quickly to rest. The supply terminals must be disconnected at suitable instant, otherwise the motor will rotate in opposite direction. If Tb is braking torque, Tl is load torque and Tpis plugging torque. If S is the slip in terms of forward rotating field, the slip in term of backward rotating field shall be 2-S. The speed of backward rotating field is(-nS) then the slip with respect to this field;

3. Where Tl=Te at a slip of S0 TP= Te at a slip of 2-S0 Note: In WRIM, the braking torque can be increased by inserting additional resistance in the rotor circuit where plugging is excited.

5. 2- The Dynamic Braking: • The plugging method suffers from the disadvantages: • Heavy losses of power is involved. • Care must be taken to de-energize the motor at a suitable instant to avoid reversal of rotation. • By injecting direct current to the stator winding requires a smaller power and is highly suitable for controlling the speed and drive during deceleration by varying the DC excitation. • The DC excitation of stator winding develops stationary field and induces voltage across the rotating rotor windings. Since the rotor winding is short circuited, current flows in it, producing a magnetic field. The magnetic field rotate as same speed of rotor but it opposite direction and it is stationary w.r.t the stator. • The magnatude of braking torque will depend on the DC voltage , rotor speed and rotor resistance.

8. 3 -AC Braking: A method of AC braking is some time used when stator winding connected with a bank of capacitor. The machine then operates as an induction generator, its excitation being supplied by the capacitors and generated energy is dissipated as a heat in the rotor winding. This method is less popular than the DC dynamic braking because of high cost of bank capacitor

9. 4 -Regenerating Braking: In this method the IM is operated as a generator and the power is returned to the supply in contrary to the dynamic braking where power generated in the rotor is dissipated in rotor copper losses. The induction generator occurs when speed of rotor is above of nS. At any instant , when the speed of rotating field is lowered below that of the rotor by increasing the stator number of poles by pole changing method. This technique is usually applied to a squirrel cage motor, when the rotor poles change simultaneously with stator poles. P 2-pole 4-pole 8-pole nS 3000rpm 1500rpm 750rpm

10. Example: A 415V, 50Hz, 8-pole, three phase IM, R1=1.5, R2=0.016, X1= 3.5, X2=0.034, effective turn ratio= 10:1, Sf =0.04. The stator is delta connected. The motor was operating on full load when plugging was executed. Calculate the braking torque immediately after plugging. Solution: nS=750rpm, R2/= 0.016 * 102=1.6, X2/=0.034 * 102=3.4

11. Dynamic of IM: Te = - mechanical torque Tm The negative (-) indicates That when Te is positive. If the speed rise Te & Tm are related as : is known as the inertia torque and its associated with the stored mechanical energy in the rotating mass. For motor mode Te is positive and Tm is negative so that; Where J= moment of inertia of the rotor and its associated mass In kg.m2. r= angular speed of rotor in rad/sec. The rotor accelerate so long as the acceleration torque Ta is positive ( Te>Tm ).

12. During the process of acceleration of the rotor, a portion of the developed torque Te balance the load torque. The remaining portion is equal to the time rate of change of mechanical energy stored in rotating mass. This time is often important as the current inrush during starting. If ( Te < Tm ) the motor will decelerate and the stored mechanical energy of the rotating mass will supply deficit torque. The motor will come to rest at a time in which the stored energy is dissipated completely.

13. For starting and plugging operation of machine , the torque equations are ; Dividing eq.1 by eq.2 and substituting Smax

14. In the whole region of motor operation, term (R1Smax) /R2/ is small compared to 1 and dominating term in the denominator:

15. is the mechanical time constant of motor. It is defined is the time taken by motor to reach its synchronous speed for standstill under constant acceleration torque equal to the maximum torque of the motor. Form eq.7 time required to start an induction motor on no load[ during starting slip change from 1 to zero, integration for S=1 to zero an infinite value is obtained, at no-load above 95% of synchronous speed ( S=0.05):

16. Thus starting time is a function of Smax. Starting time has a minimum value of at Smax = 0.4. The rotor resistance required to start the motor in minimum time is ( neglecting R1):

17. The rotor winding loss for starting can be written as:

18. Sub.eq.14 in eq.13; The rotor winding energy loss is equal to the kinetic energy stored in moving parts at completion of the starting process and it is independent of the starting time or rotor resistance. Energy loss in stator winding( neglecting magnetizing current) is; Total winding loss during starting at no-load; The rotor winding loss during stopping by plugging under no-load; Energy dissipated in the machine during braking operation;

19. Example: A 2200V, 50Hz, 3-phase, 6-pole, Y-connected squirrel cage IM has the following parameters: R1=0.075X1= 0.5 R2/=0.12X2/=0.5 The combined inertia of motor and load is 100kg.m2 . 1-calculate time taken and energy dissipated in the motor during starting. 2-Calculate time taken and energy dissipated in the motor when its stopped by plugging. 3-What resistance should inserted in the rotor to stop motor by plugging in the minimum time?

20. The energy dissipated in the motor; = 891KJoule 3- From eq.10 the minimum time; Must be inserted to decreasing the stopping time from 3Sec to 0.5 Sec.