ROTARY CONVERTERS 103 



on the alternating-current side of its full-load value? Neglect 

 any change in the efficiency. 



56. ' At full load and when operating at unit power factor a 

 6-phase, 60-cycle rotary converter takes a line current of 840 

 amperes at 212 volts per phase and delivers 1000 kw. at 600 

 volts. The armature has 180 slots with 6 inductors in series 

 per slot. The field structure has 12 poles with 864 turns in the 

 shunt winding and 2 turns in the series winding per pole. The 

 resistance of the shunt field circuit is 39.65 ohms with the regu- 

 lating rheostat cut out. A current of 9.3 amperes in the shunt 

 field alone is the normal excitation, i.e., for full load and unit 

 power factor. With an output of 800 kw. at 600 volts what is 

 the least power factor at which the rotary can be operated when 

 overexcited? What per cent, is the current of its full-load value? 

 Neglect any change in this efficiency. 



57. ! A 3-phase, 25-cycle rotary converter has a full-load capa- 

 city of 300 kw. at 600 volts. The armature has 96 slots with 

 6 inductors in series per slot. The field structure has 4 poles 

 with 2340 turns in the shunt winding per pole. The resistance 

 of this field circuit with all of the regulating rheostat cut out is 

 120.5 ohms. A current of 3.25 amperes in the shunt field winding 

 alone gives the normal excitation, i.e., for full load at unit power 

 factor. In order to maintain the terminal voltage on the direct- 

 current side at 600 volts when full load is delivered, it is neces- 

 sary to overexcite the rotary so that it takes a line current of 

 600 amperes at a power factor of 86 per cent. 



Calculate the least number of series field turns that must be 

 added in order that this excitation may be produced. What 

 should be the current in the shunt field winding? 



58. l Calculate the efficiency of the rotary A when it receives 300 

 kw. at a power factor of 0.90 and with a lint- voltage of 380 

 volts. The excitation is greater than normal. Assume that 

 the running temperature is 70 C. 



59. 1 Calculate the efficiency of the rotary A wlu-n it <1 livers ;i()() 

 kw. at 625 volts and the excitation ispn-at.-!- than normal and u 

 adjusted so that the line current on the alternating-current side 

 is 520 amperes. Assume that the running temperature is 70 C. 



1 ID ralculat inn flu- armature reaction do not HIT!- t!,.- distribution of 

 tin- winding, and assume that the constant usually mv -n as 0.707 is 0.60. 

 This makes an approximate correction for th< < tTrct of the ratio of pole arc 

 to pole pitch. As a first approximation it may be assumed that the effi- 

 ciency is 95 per cent. 



