Skip to main content

Slip of Induction Motor

When have seen that rotor rotates in the same direction as that of R.M.F. but in steady state attains a speed less than the synchronous speed. The difference between the two speeds i.e. synchronous speed of R.M.F. ( Ns ) and rotor speed (N) is called slip speed. This slip speed is generally expressed as the percentage of the synchronous speed.
       So slip of the induction motor is defined as the difference between the synchronous speed ( Ns) and actual speed of rotor i.e. motor (N) expressed as a friction of the synchronous speed ( Ns ). This is also called absolute slip or fractional slip and is denoted as 's'.
        Thus 


       The percentage slip is expressed as,

       In terms of slip, the actual speed of motor (N) can be expressed as,

        At start, motor is at rest and hence its speed N is zero.

       This is maximum value of slip s possible for induction motor which occurs at start. While s = 0 given us N = Ns which is not possible for an induction motor. So slip of induction motor can not be zero under any circumstances.
       Practically motor operates in the slip range of 0.01 to 0.05 i.e. 1 % to 5 %. The slip corresponding to full load speed of th motor is called full load slip.
Example 1 : A 4 pole, 3 phase induction motor is supplied from Hz supply. Determine its synchronous speed. On full load, its speed is observed to be 1410  r.p.m. calculate its full load slip.
Solution : Given values are,
                              P = 4,            f = 50 Hz ,               N = 1410 r.p.m.
                             Ns = 120f / P = 120 x 50 / 4 = 1500 r.p.m.
       Full load absolute slip is given by,
                             s = ( Ns - N)/ V2 = (1500-1410 )/ 1500 = 0.06
...                            %s = 0.06 x 100 = 6 %

Example 2 : A 4 pole, 3 phase, 50 Hz, star connected induction motor has a full load slip of 4 %. Calculate full load speed of the motor.
Solution : Given values are,
                                 P = 4,           f = 50 Hz,                % sfl = 4%
                                 sfl = Full load absolute slip = 0.04
                                  Ns = 120f  / P = 120 x 50 / 4 = 1500 r.p.m.
                                   sfl = (Ns - Nfl ) / Ns =  where   =  full load speed of motor
...                                 0.04 = (1500 - Nfl )/ 1500
...                                  Nfl = 1440 r.p.m.

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 a...

Armature Voltage Control Method or Rheostatic Control of dc motor

Speed Control of D.C. Shunt Motor (Part2)  2. Armature Voltage Control Method or Rheostatic Control        The speed is directly proportional to the voltage applied across the armature. As the supply voltage is normally constant, the voltage across the armature can be controlled by adding a variable resistance in series with the armature as shown in the Fig. 1. Fig. 1 Rheostat control of shunt motor        The field winding is excited by the normal voltage hence I sh is rated and constant in this method. Initially the reheostat position is minimum and rated voltage gets applied across the armature. So speed is also rated. For a given load, armature current is fixed. So when extra resistance is added in the armature circuit, I a remains same and there is voltage drop across the resistance added ( I a R). Hence voltage across the armature decreases, decreasing the speed below normal value. By varyi...