Page 302 - Solutions Manual to accompany Electric Machinery Fundamentals
P. 302

B-2.   Derive the relationship for the winding distribution factor  k d   in Equation B-22.



































                 SOLUTION  The above illustration shows the case of 5 slots per phase, but the results are general.  If there
                 are 5 slots per phase, each with voltage  E Ai , where the phase angle of each voltage increases by  from
                 slot to slot, then the total voltage in the phase will be
                         E   E    E A2    E A3    E A4    E A5   ...   E An
                           A
                                A1
                      The resulting voltage  E A  can be found from geometrical considerations.  These “n” phases, when
                 drawn end-to-end, form equally-spaced chords on a circle of radius R.  If a line is drawn from the center
                 of a chord to the origin of the circle, it forma a right triangle with the radius at the end of the chord (see
                 voltage  E A 5  above).  The hypotenuse of this right triangle is R, its opposite side is  E  2 /  , and its smaller
                 angle is  /  2 .  Therefore,

                                                    1
                                E 2/                E
                         sin                 R   2                                                   (1)
                            2     R                   
                                                   sin
                                                      2
                 The total voltage  E A   also forms a chord on the circle, and dropping a line from the center of that chord to
                 the origin forms a right triangle.  For this triangle, the hypotenuse is R, the opposite side is  E A  2 /  , and
                 the angle is  n  2 / .  Therefore,
                                                           1
                            n    E / 2                      E A
                         sin      A                 R   2                                            (2)
                             2      R                          n
                                                          sin
                                                              2

                 Combining (1) and (2) yields



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