Page 74 - Solutions Manual to accompany Electric Machinery Fundamentals
P. 74

(b)  The autotransformer connection for 600/480 V stepdown operation is
                                                +            +

                                                        N     V
                                                         SE     SE
                                                              -
                                              600 V          +         +
                                                        N    V
                                                         C     C
                                                                     480 V

                                                 -            -        -
                 (c)  When used as an autotransformer, the kVA rating of this transformer becomes:

                            N   N       41
                                          
                       S   IO  C  SE  S   W  10 kVA     50 kVA
                              N SE        1
                 (d)  As an autotransformer, the per-unit series impedance  Z EQ  is decreased by the reciprocal  of  the

                 power advantage, so the series impedance becomes
                            0.00963
                       R             0.00193 pu
                        EQ
                               5
                             0.0398
                       X            0.00796 pu
                        EQ
                               5
                 while the magnetization branch elements are basically unchanged.  At rated conditions and unity power
                 factor, the output power to this transformer would be  P OUT  = 1.0 pu.  The input voltage would be


                          V    V  I  Z
                        P    S    S  EQ
                           1 0  V 
                       V P       1 0  0.00193  j 0.00796   1.002 0.5  pu    
                 The core losses (in resistor  R C  ) would be

                              2  V    1.002  2
                       P                 0.00382 pu
                        core
                            R C    263
                 The copper losses (in resistor  R EQ ) would be

                             2
                       P   CU  I R EQ     1.0  2   0.00193     0.0019 pu
                 The input power of the transformer would be

                                               
                                                       
                                                                
                       P   IN  P OUT    P   CU  P core    1.0 0.0019 0.00382 1.0057
                 and the transformer efficiency would be
                          P             1.0
                         OUT    100%       100% 99.4%
                                                    
                           P IN        1.0057
                 The voltage regulation of the transformer is

                                  
                            1.0057 1.00
                       VR               100% 0.6%
                                              
                                1.00

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