Page 297 - Solutions Manual to accompany Electric Machinery Fundamentals
P. 297

(e)  The real and reactive power supplied by the generator when the switch is closed is just the sum of
                 the real and reactive powers of Loads 1, 2, and 3.  The powers of Loads 1 and 2 have already been
                 calculated.  The real and reactive power of Load 3 are:
                         P   80 kW
                          3
                                              
                         Q   P tan   P tan  cos PF  80 kW tan      31.79           49.6 kvar
                                               1
                          3                                              
                                                         
                         P   P   P   P   100 kW 64 kW 80 kW   244 kW
                                                 
                          G   1   2   3
                                                            
                                                    
                         Q   G  Q   1  Q   2  Q   3  48.4 kvar 48 kvar 49.6 kvar   46.8 kvar
                                                                               P                     Q
                 (f)   The line current when the switch is closed is given by  I     , where    tan  1  G  .
                                                                        L
                                                                            3 V  cos                 P
                                                                                L                      G
                                 Q         46.8 kvar
                            tan   1  G    tan   1    10.86
                                 P G        244 kW
                                  P              244 kW
                         I                                     298.8 A
                          L
                               3 V L  cos       V cos      10.86
                                                    3 480
                 (g)  The total line current from the generator is 298.8 A.  The line currents to each individual load are:
                                  P            100 kW
                         I        1                       133.6 A
                                                     
                           1
                          L
                               3 V L  cos 1        3 480 V 0.9 
                                S       80 kVA
                         I      2                96.2 A
                          L
                           2
                               3 V L    3    V 480
                                   P            80 kW
                         I L 3     3                        113.2 A
                                                      
                               3 V L  cos 3         3 480 V 0.85 
                 The sum of the three individual line currents is 343 A, while the current supplied by the generator is 298.8
                 A.  These values are not the same, because the three loads have different impedance angles.  Essentially,
                 Load 3 is supplying some of the reactive power being consumed by Loads 1 and 2, so that it does not
                 have to come from the generator.
          A-4.   Prove that the line voltage of a Y-connected generator with an acb phase sequence lags the corresponding
                 phase voltage by 30.  Draw a phasor diagram showing the phase and line voltages for this generator.
                 SOLUTION  If the generator has an acb phase sequence, then the three phase voltages will be
                                   V  V   0 
                                an  
                                   V  V     240 
                                bn  
                                   V  V     120 
                                cn  
                 The relationship between line voltage and phase voltage is derived below.  By Kirchhoff’s voltage law,
                 the line-to-line voltage  V ab   is given by
                                   V    V  V
                                 ab   a    b
                                    V  V   0     V     240 
                                 ab          
                                           1       3     3        3
                                    V  V     V   j  V    V   j  V
                                 ab                             
                                           2       2     2       2
                                            3   1 
                                V     3V       j
                                 ab              
                                            2   2
                                               
                                    V  3V     30
                                 ab      

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