Page 114 - Electrician - TT (Volume 2)
P. 114

ELECTRICIAN - CITS



           No-load test of induction motors

           The induction motor is connected to the supply through a three phase auto transformer, the three phase auto
           transformer is use to regulate the starting current by applying low voltage at the start and the gradually increased
           to rated voltage. The ammeter and volt meters are selected based upon the motor specification. The no-load
           current of the motor will be very low up to 30% of full load.
           The power factor of the motor on no load is very low in the range of 0.1 to 0.2 the watt meters selected are such
           as to give a current reading at low power factor. The wattmeter full scale reading will approximate equal to the
           product of the ammeter and voltmeter full scale deflection values

           At no load the outfit delivered by the motor is zero.  All the mechanical power developed in the rotor is used to
           maintain the rotor running at its rated speed. Hence the input power is equal to the no-load copper loss plus iron
           losses and mechanical losses.

           Slip ring induction motor are started with full line voltage across the stator winding. To reduce the heavy starting
           current a star connected external resistance is added in the rotor circuit. The external resistances are cut out and
           the rotor winding ends are shorted once the motor picks up speed. Now days semi-automatic starter is used. By
           pressing the on button the contactor will close only when the shorting point A at the rotor resistance is in closed
           position. This is possible only when the handle is in the start position once the motor starts running the handle of
           the rotor resistance should be brought to run position to cut-out the rotor resistance.
           V = line stator voltage
            NL
           I  = line current
            NL
           P = power input
            Nl
           The input power consists of the core loss Pc, friction and windage loss P(rot)  and stator copper loss
           Pnl = Pc+Prot+3I Rs
                          2
           Where stator resistance Rs per phase obtained from a resistance measurement at the stator terminal
           In star connection Rs =R/2
           Delta connection Rs =2/3R
           Blocked rotor rest
           The connections are made similar to that of the no-load test. In this case the ammeter is selected to carry the full
           load current of the motor. Wattmeterswill be of a suitable range and its power factor is 0.5 to unity.
           An auto transformer is used to giber a much lower percentage of the rated voltage. The rotor is locked by a
           suitable arrangement such that it cannot rotate even if the supply is given to the motor. The belt is over tightened
           on the pulley to prevent rotation.
           As the rotor is in a locked condition it is equivalent to the short circuit secondary of a transformer. The   frequency
           of the starter supply voltage is maintained at normal rated supply frequency.
           The method of calculating the copper losses from the result is illustrated through the example given below.
           Example
           A 5 hp 400 V, 50 HZ four pole three phase induction motor was tested and the following data were obtained.

           Blocked rotor test V  =54, PS=430, I =7.5A.The resistance of the stator winding gives a 4V drop between the
                             S
                                            S
           terminals rated DC current flowing.
           W  = √3 Vs Is COS Фs
             S
                           Vs
           Impedance Ze =   √3Is   = √Re  +Xe
                                          2
                                      2
                      Ws
           Cos Ф =   √3 Vs Is
                       = 430/ (1.72*54*7.5) = 0.61
           Equivalent resistance Re /phase = Ps/(3xIs )  = 430/(3*7.52 ) = 2.5Ω
                                                 2
           Equivalent reactance /phase =√Ze - Re 2
                                         2


                                                           101

 CITS : Power - Electrician & Wireman - Lesson 70-75  CITS : Power - Electrician & Wireman - Lesson 70-75
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