38 PROBLEMS IN ALTERNATING CURRENT MACHINERY 



live load). What is the field current calculated by this method? 

 Assume that the temperature of the armature windings is 

 70 C. 



(b) Calculate the regulation of this generator for the specified 

 load by the synchronous impedance method. What is the field 

 current calculated by this method? 



(c) Calculate the regulation of this generator for the specified 

 load by the magnetomotive-force method. What is the field 

 current calculated by this method? 



56. A 2-phase, 60-cycle alternating-current generator is rated 

 to deliver 100 kv.-a. at 480 volts. The armature has an effective 

 resistance of 0.138 ohm and a leakage reactance of 0.159 ohm per 

 phase. When the power factor of the load is 0.8 the armature 

 demagnetizing ampere turns are 11.8 and the cross-magnetizing 

 ampere turns are 15.2 per pole per ampere. The field poles 

 are each wound with 265 turns. The data for the open circuit 

 characteristic are: 



What is the regulation of this generator when delivering full- 

 load current at 0.8 power factor (inductive load) ? 



(a) Assume that the cross-magnetizing and demagnetizing 

 ampere turns act on magnetic circuits of the same reluctance as 

 that of the resultant field. This is a modification of the general 

 method. 



(b) Assume that the cross-magnetizing ampere turns act on a 

 magnetic circuit whose reluctance is determined by the lower 

 part of the saturation curve, and that the demagnetizing ampere 

 turns act on the same magnetic circuit as do the impressed field 

 ampere turns. This is the Blondel method. 



57. A 3-phase, 25-cycle alternating-current generator is rated 

 to deliver 850 kv.-a. at 5000 volts. The armature windings are 

 connected in Y, and have an effective resistance of 0.398 ohm and 

 a leakage reactance of 1.46 ohms per phase. When the power 

 factor of the load is unity the armature demagnetizing ampere 

 turns are 5.8 and the cross-magnetizing ampere turns are 31 per 



