Page 113 - Electrician - TT (Volume 2)
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ELECTRICIAN - CITS




           Losses and efficiency of induction motor
           There are two types of losses occur in three phase induction motor. These losses are,
           1  Constant or fixed losses,
           2  Variable losses.

           Constant or fixed losses
           Constant losses are those losses which are considered to remain constant over normal working range of induction
           motor. The fixed losses can be easily obtained by performing no-load test on the three phase induction motor.
           These losses are further classified as-
           1  Iron or core losses,
           2  Mechanical losses,
           3  Brush friction losses

           Iron or core losses
           Iron or core losses are further divided into hysteresis and eddy current losses. Eddy current losses are minimized
           by using lamination. Since by laminating the core, area decreases and hence resistance increases, which results
           in decrease in eddy currents. Hysteresis losses are minimized by using high grade silicon steel. The core losses
           depend upon frequency. The frequency of stator is always supply frequency, f and the frequency of rotor is slip
           times the supply frequency, (sf) which is always less than the stator frequency. Hence the rotor core loss is very
           small as compared to stator core loss and is usually neglected in running conditions.

           Mechanical and brush friction losses
           Mechanical losses occur at the bearing and brush friction loss occurs in wound rotor induction motor. These
           losses occur with the change in speed. In three phase induction motor the speed usually remains constant. hence
           these losses almost remain constant.
           Variable losses
           These losses are also called copper losses. These losses occur due to current flowing in stator and rotor winding.
           As the load changes, the current flowing in rotor and stator winding also changes and hence these losses also
           changes. Therefore,  these  losses  are  called  variable  losses. The  copper  losses  are  obtained  by  performing
           blocked rotor test on three phase induction motor.
           The  induction motor electrical power the input to the three phase induction motor is three phase supply. So, the
           three phase supply is given to the stator of three phase induction motor. Let, P  = electrical power supplied to the
                                                                                in
           stator of three phase induction motor, V  = line voltage supplied to the stator of three phase induction motor, I
                                               L
                                                                                                             L
           = line current, Cosφ = power factor of the three phase induction motor. Electrical power input to the stator, P  =
                                                                                                           in
           √3V I cosφ A part of this power input is used to supply stator losses which are stator iron loss and stator copper
               L L
           loss. The remaining power i.e( inputelectrical power – stator losses) are supplied to rotor as rotor input. So, rotor
           input P  = P  – stator losses (stator copper loss and stator iron loss).
                      in
                 2
           Now, the rotor has to convert this rotor input into mechanical energy but this complete input cannot be converted
           into mechanical output as it has to supply rotor losses. As explained earlier the rotor losses are of two types rotor
           iron loss and rotor copper loss. Since the iron loss depends upon the rotor frequency, which is very small when
           the rotor rotates, so it is usually neglected. So, the rotor has only rotor copper loss. Therefore, the rotor input has
           to supply these rotor copper losses. After supplying the rotor copper losses, the remaining part of Rotor input, P 2
           is converted into mechanical power, P .
                                             m
           Let P  be the rotor copper loss,
                c
           I  be the rotor current under running condition,
            2
           R  is the rotor resistance,
            2
           P  is the gross mechanical power developed.
            m
           P  = 3I R 2
                  2
            c
                 2



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