Page 151 - Solutions Manual to accompany Electric Machinery Fundamentals
P. 151

  XP            0.70 0.748     
                            sin     1  S      sin     1        8.4
                                   3VE A        3  1.0 1.20         
                                    

                 (The negative sign on  comes from the fact that  E   lags  V  in a motor.)
                                                              A       
                 At rated voltage and 12 MW of power supplied, the armature current will be

                                         
                              V   E    1 0  1.20  8.4
                                                                
                                                                      
                         I        A                      0.366 46.9  pu
                          A
                             R   jX S        j 0.70
                               A
                 In actual amps, this current is
                         I A    1404 A 0.396 46.9     556 46.9  A    
                 The reactive power supplied at the conditions of maximum  E A   and 12 MW power is

                                                 
                         Q   3V I A      sin    3 3811 V 556 A sin 46.9       4.6 MVAR
                               
                 Therefore, the field current limit occurs before the stator current limit for these conditions, and the
                 maximum reactive power that the motor can supply is 4.6 MVAR under these conditions.
                 (d)  At rated conditions, the field circuit consumes

                         P field    V I   FF  125 V 5.2 A     650 W
                 (e)  The efficiency of this motor at full load is

                             P                    21000 hp  746 W/hp 
                            OUT    100%                      100%   97.6%
                             P IN              16.05 MW

                 (f)   The output torque in SI and English units is
                                          P      21000 hp  746 W/hp 
                                                                             
                              OUT                                124,700 N m
                          load
                                m             1200 r/min        1 min  60 s           2  rad   
                                                           1 r 
                                             
                               5252 P   5252 21000 hp 
                                                       91,910 lb ft 
                          load
                                  m        n       1200 r/min
          5-16.  A 480-V, 500-kVA, 0.8-PF-lagging, Y-connected synchronous generator has a synchronous reactance of
                 0.4  and a negligible armature resistance.  This generator is supplying power to a 480-V, 80-kW, 0.8-
                 PF-leading,  Y-connected synchronous motor with a synchronous reactance of 2.0   and a negligible
                 armature resistance.  The synchronous generator is adjusted to have a terminal voltage of 480 V when the
                 motor is drawing the rated power at unity power factor.
                     (a)  Calculate the magnitudes and angles of E  A  for both machines.
                     (b)  If the flux of the motor is increased by 10 percent, what happens to the terminal voltage of the
                     power system?  What is its new value?
                     (c)  What is the power factor of the motor after the increase in motor flux?

                 SOLUTION

                 (a)  The motor is Y-connected, and is operating at rated voltage (480 V), rated power (80 kW), and unity
                 power factor.  The current flowing in the motor is




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