Page 427 - Handbook of Electrical Engineering
P. 427

416    HANDBOOK OF ELECTRICAL ENGINEERING

                    The average value of the waveform is zero because it is symmetrical about the Y-axis, and so
              the coefficient a o for the average value is zero. The sinusoidal function in the coefficient a n varies
              with the commutation angle u and approaches a limiting value when u is small,

                                                            un
                                                        2sin
                                                             2     1
                                            As u → 0,           →
                                                          un 2     n
                                                                      ◦
                    Which applies to a rectangular waveform. When u is 60 the sinusoidal function has an
              absolute value of,

                                                          un
                                                      2sin
                                                                 3

                                                  ◦        2     =
                                                       un 2     πn 2
                                            u = 60 ,


                                                0.9549
                                              =
                                                  n 2
                    Therefore the magnitude of all the harmonics decrease as u increases, which is a reasonable
              expectation since the waveform more closely resembles a sine wave.
                                                                             √
                    The magnitude of the sum of the four cosine terms in (15.18) is 2 3 for all values of k in
              (15.19), otherwise the magnitude is zero.
                    Table 15.2 shows the magnitudes of b n after scaling them by 1/b 1 ,i.e.creating b 1 = 1.0
              as reference.


              15.3.3.1 Worked example
              Consider a 250 kW DC motor fed by a rectifier system. The line voltage is 415 volts at 50 Hz. The
              rectifier is fed by a 400 kVA transformer which has an unusually high impedance of 0.0 + 24.5%.
              Assume the motor rated efficiency is 0.9 per unit. Assume the motor terminal voltage is 262.3 volts
              and its total current is 425 amps.
                                                415
                    Phase voltage of the supply E = √ = 239.6 volts.
                                                  3
                                                             √
                                                            3 6
                    Open-circuit DC voltage of the rectifier V do =  (239.6) = 560.45 volts.
                                                             π
                    The supply current


                                               2I d  3
                                         I ac =      = 0.7797 I d
                                               π    2
                                            = 0.7797 × 425 = 331.37 amps

                                           400,000
              The transformer rated current = √    = 556.48 amps
                                            3 × 415
                               239.6
              1 pu impedance =       = 0.4306 ohms/phase
                              556.48
   422   423   424   425   426   427   428   429   430   431   432