Page 61 - Solutions Manual to accompany Electric Machinery Fundamentals
P. 61

  1       1 
                       I        I               27.57 A   2.757 A
                        line      load    
                              10      10 
                 and the losses in the transmission line are

                       P loss    line 2 R line     I  2.757 A  2   30        228 W
                 (d)  The efficiency of this power system is

                           P               P                    190 kW
                                                                                     
                          out    100%   out    100%                       100% 99.9%
                                                                  
                           P in         P   out  P loss  190 kW 0.228 kW
                 (e)  Transmission losses have decreased by a factor of more than 80 when the transformers were added
                 to the system.

          2-15.  An autotransformer is used to connect a 12.6-kV distribution line to a 13.8-kV distribution line.  It must
                 be capable of handling 2000 kVA.  There are three phases, connected Y-Y with their neutrals solidly
                 grounded.
                 (a) What must the  N /  N SE  turns ratio be to accomplish this connection?
                                      C
                 (b)  How much apparent power must the windings of each autotransformer handle?
                 (c)  What is the power advantage of this autotransformer system?
                 (d)  If one of the autotransformers were reconnected as an ordinary transformer, what would its ratings
                      be?
                 SOLUTION

                 (a)  The  transformer  is  connected Y-Y, so the primary and secondary phase voltages are the line
                 voltages divided by  3 .  The turns ratio of each autotransformer is given by

                       V H    N   N SE    13.8 kV/ 3
                             C
                       V L     N C    12.6 kV/ 3
                      12.6 N   C  12.6 N SE    13.8 N
                                               C
                      12.6 N SE   1.2 N
                                     C
                                             
                 Therefore, N C  / N SE    12.6 /1.2 10.5 .
                 (b)  The power advantage of this autotransformer is

                       S IO    N   C  N SE    N   C  10.5N C    11.5
                       S W     N C         N C
                 so 1/11.5 of the power in each transformer goes through the windings.  Since 1/3 of the total power is
                 associated with each phase, the windings in each autotransformer must handle
                            2000 kVA
                       S               63.5 kVA
                        W
                            10.5 
                             3
                 (c)  As determined in (b), the power advantage of this autotransformer system is 11.5.

                 (d)  The voltages across each phase of the autotransformer are 13.8/ 3  = 7967 V and 12.6 / 3  = 7275
                 V.  The voltage across the common winding ( N C  ) is 7275 kV, and the voltage across the series winding
                 ( N SE  ) is 7967 kV – 7275 kV = 692 V.  Therefore, a single phase of the autotransformer connected as an
                 ordinary transformer would be rated at 7275/692 V and 63.5 kVA.

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