Page 438 - Handbook of Electrical Engineering
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HARMONIC VOLTAGES AND CURRENTS       427

           Table 15.4. Harmonic rms currents and voltages in a star-wound induction motor that is fed from a voltage
           source inverter
           Harmonic      Stator current     Rotor current      Magnetising       Air-gap voltage
           number                                                current

                       Mag.     Angle     Mag.     Angle     Mag.     Angle     Mag.     Angle
                      (Amps)  (Degrees)  (Amps)   (Degrees)  (Amps)  (Degrees)  (Amps)  (Degrees)
               1      392.28   −25.60    371.51   −14.27     78.56   −89.07    230.27   −3.93
               5       62.14   −89.35     59.86   −89.33      2.27   −89.72     33.32     0.28
               7       31.70   −89.46     30.54   −89.45      1.16   −89.83     23.80     0.17
              11       12.84   −89.69     12.37   −89.69      0.47   −89.88     15.14     0.12
              13        9.19   −89.72      8.86   −89.71      0.37   −89.90     12.81     0.09
              17        5.38   −89.80      5.18   −89.79      0.20   −89.92      9.80     0.08
              19        4.30   −89.81      4.14   −89.80      0.16   −89.93      8.77     0.07
              23        2.94   −89.85      2.83   −89.85      0.11   −89.94      7.24     0.06
              25        2.49   −89.86      2.40   −89.85      0.09   −89.95      6.66     0.05
              29        1.85   −89.88      1.78   −89.88      0.07   −89.96      5.74     0.05
              31        1.62   −89.89      1.56   −89.88      0.06   −89.96      5.37     0.04



                 If the above calculations are made for all the active harmonics then their results can be added
           and the waveforms synthesised. Table 15.4 summarises the results.
                 Figure 15.12 shows the synthesised currents and air-gap voltage using the first 61 harmonics.



           15.4.3.2 Worked example

           The same motor as used in the ‘worked example’ of sub-section 15.4.3.1 is fed from a current source
           inverter. Find the currents and air-gap voltage in the circuit.
                 The equivalent circuit fed from a constant current source is shown in Figure 15.11, wherein
           I 1n is the source current instead of V 1n .
                                                                    ◦
                 Assume the inverter output line current consists of a 120 rectangle wave and that the
           commutation angle u is small enough to be ignored. The complete waveform has a harmonic con-
           tent of,

                                   1      πn       5πn       7πn      11πn
                              b n =    cos    − cos    − cos     + cos
                                   πn      6        6         6         6
                                   3.464
                                =         for n = 1, 5, 7, 11, 13 etc.
                                    nπ
                                   1.1026
                                =
                                     n

                 The rms value of the fundamental line current is 392.26 amps. Therefore its peak value is
                 √
           392.26 2 = 554.74 amps which corresponds to b 1 having a value of 1.1026. The peak values of the
           harmonic components of the line current are given below in Table 15.5.
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