Page 306 - Solutions Manual to accompany Electric Machinery Fundamentals
P. 306

. 0  342
                       th
                      7 :                . 0  347
                                 . 0  985
                                                                                                     th
                                    th
                 In other words, the 5  harmonic is suppressed by 34.7% relative to the fundamental, and the 7  harmonic
                 is suppressed by 65.3% relative to the fundamental frequency.
          B-7.   What coil pitch could be used to completely eliminate the seventh-harmonic component of voltage in ac
                 machine armature (stator)?  What is the minimum number of slots needed on an eight-pole winding to
                 exactly achieve this pitch?  What would this pitch do to the fifth-harmonic component of voltage?
                 SOLUTION  To totally eliminate the seventh harmonic of voltage in an ac machine armature, the pitch factor
                 for that harmonic must be zero.
                                     7
                         k    0   sin
                          p
                                      2
                         7
                                 180  n   ,    n = 0, 1, 2, …
                          2
                             2     180   n
                          
                                 7
                 In  order  to maximize the fundamental voltage while canceling out the seventh harmonic, we pick the
                 value of n that makes  as nearly 180 as possible.  If n = 3, then  = 154.3, and the pitch factor for the
                 fundamental frequency would be
                                 154 3 . 
                         k    sin        . 0  975
                          p
                                    2
                 This pitch corresponds to a ratio of 6/7.  For a two-pole machine, a ratio of 6/7 could be implemented
                 with a total of 14 slots.  If that ratio is desired in an 8-pole machine, then 56 slots would be needed.

                 The fifth harmonic would be suppressed by this winding as follows:
                                 154 5  3 .   
                         k    sin            . 0  434
                          p
                                      2
          B-8.   A 13.8-kV Y-connected 60-Hz 12-pole three-phase synchronous  generator has 180 stator slots with a
                 double-layer winding and eight turns per coil.  The coil pitch on the stator is 12 slots.  The conductors
                 from all phase belts (or groups) in a given phase are connected in series.
                 (a) What flux per pole would be required to give a no-load terminal (line) voltage of 13.8 kV?
                 (b) What is this machine’s winding factor  k w  ?

                 SOLUTION

                 (a)  The stator pitch is 12/15 = 4/5, so    144 , and
                                 144
                         k    sin      . 0  951
                          p
                                   2
                 Each phase belt consists of (180 slots)/(12 poles)(6) = 2.5 slots per phase group.  The slot pitch is 2
                 mechanical degrees or 24 electrical degrees.  The corresponding distribution factor is
                                    n        5   . 2   24
                              sin      sin
                         k d     2          2 24    . 0  962
                                 n sin   2    .5 sin
                                   2            2

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