Page 54 - Solutions Manual to accompany Electric Machinery Fundamentals
P. 54

V    510 V              I SC   12  A   6 .      P   3000 W
                                                                     SC
                        SC
                 (a)  If this bank delivers a rated load at 0.8 PF lagging and rated voltage, what is the line-to-line voltage
                      on the primary of the transformer bank?
                 (b)  What is the voltage regulation under these conditions?

                 (c)  Assume that the primary phase voltage of this transformer is a constant 8314  V,  and  plot  the
                      secondary voltage as a function of load current for currents from no-load to full-load.  Repeat this
                      process for power factors of 0.8 lagging, 1.0, and 0.8 leading.
                 (d)  Plot the voltage regulation of this transformer as a function of load current for currents from no-load
                      to full-load.  Repeat this process for power factors of 0.8 lagging, 1.0, and 0.8 leading.
                 (e)  Sketch the per-unit equivalent circuit of this transformer.

                 SOLUTION  From the short-circuit information, it is possible to determine the per-phase impedance of the
                 transformer bank referred to the high-voltage (primary) side.  Note that the short-circuit information is
                 given for one transformer of the three in the bank.  The voltage across this transformer is

                      V  ,SC   510 V

                 the short-circuit phase current is
                       I  ,SC   12.6 A

                 and the power per phase is
                       P  ,SC    3000 W

                 Thus the per-phase impedance is

                                          510 V
                                                        
                       Z     R    jX           40.48
                                     EQ
                        EQ
                               EQ
                                          12.6 A
                                P              3000 W
                         cos   1  SC    cos   1         62.1
                                                   
                               VI                510 V 12.6 A 
                                SC SC
                                             
                       Z EQ    R EQ    jX EQ    40.48 62.1       18.94   j   35.77
                       R EQ    18.94  
                       X EQ    j 35.77  
                 (a)  If this Y- transformer bank delivers rated kVA (300 kVA) at 0.8 power factor lagging while the
                 secondary voltage is at rated value, then each transformer delivers 100 kVA at a voltage of 480 V and 0.8
                 PF lagging.  Referred to the primary side of one of the transformers, the load on each transformer is
                 equivalent to 100 kVA at 8314 V and 0.8 PF lagging.  The equivalent current flowing in the secondary of
                 one transformer referred to the primary side is
                           100 kVA
                       I              12.03 A
                       
                        ,S
                             8314 V
                         
                                          
                         ,S   I  12.03    36.87  A
                 The voltage on the primary side of a single transformer is thus
                                      
                       V  P ,    V  S ,    I  S ,   Z EQ, P

                             8413 0  V      V    36.87  A 12.03  18.94   j 35.77     8856 1.34  V    
                                                       
                                 
                         ,P
                 The line-to-line voltage on the primary of the transformer is
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