Page 115 - Solutions Manual to accompany Electric Machinery Fundamentals
P. 115

SOLUTION  If the load is suddenly removed, V  =  E A .  The internal generated voltage  E A   at full load was
                                                         
                 2814 V, so  V  = 2814 V after the load is removed.  Since  the  generator  is  Y-connected,
                                
                 V     3V   4874 V  when the load is removed.
                   T      
          4-20.  What are the electrical losses in this generator at rated conditions?
                 SOLUTION  The current flow at rated conditions is  I   90.2 A , so the electrical losses are
                                                              A
                         P   3I  2      R    3 90.2 A  2    0.7     17.1 kW
                          CU    A  A
          4-21.  If this machine is operating at rated conditions, what input torque must be applied to the shaft of this
                 generator?  Express your answer both in newton-meters and in pound-feet.

                 SOLUTION  To get the applied torque, we must know the input power.  The input power to this generator is
                 equal to the output power plus losses.  The rated output power and the losses are
                                      
                         P     1 MVA 0.9  900 kW
                          OUT
                         P   3I  2      R    3 90.2 A  2    0.7     17.1 kW
                          CU    A  A
                         P     10 kW
                          F&W
                         P core   8 kW
                         P     (assumed 0)
                          stray
                         P   P     P   P     P    P    935.1 kW
                          IN   OUT   CU   F&W   core  stray
                 Therefore, the applied torque is

                               P               935.1 kW
                              IN                                   7441 N m
                                                                             
                          APP
                                m              1200 r/min    2 rad       1 min 
                                                      1 r          60 s   
                                              
                               7.04 P   7.04 935.1 kW   
                 or       APP                            5486 lb ft 
                                 n m       1200 r/min

          4-22.  What is the torque angle  of this generator at rated conditions?
                 SOLUTION  From the calculations in Problem 4-17,    26.3 .

          4-23.  Assume that the generator field current is adjusted to supply 3200 V under rated conditions.  What is the
                 static stability limit of this generator?  (Note: You may ignore  R A  to make this calculation easier.)  How
                 close is the full-load condition of this generator to the static stability limit?
                 SOLUTION  From Problem 4-17, the phase voltage of this generator at rated conditions is  V    1850 0
                                                                                      
                                                                                 
                 and the internal generated voltage  at rated conditions  is  E    2814 26.3  V .    Therefore, the static
                                                                         A
                 stability limit is
                                                  
                               3 VE      3     1850 V 2814 V 
                         P          A                      995 kW
                          MAX
                                  X  S         15.7 
                 The maximum possible power of this generator is about twice the full-load rated power.



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