CHAP. VIII] REACTION IN SYNCHRONOUS MACHINES 157 



Since <' > 90 degrees, the angle ft and the voltage E t are negative. 

 The angle <f> z also is usually negative. The cases of a leading 

 current in the generator and of a lagging current in the motor are 

 obtained by assigning the proper value and sign to the angle <. 

 For the application of the Blondel diagram to the determination 

 of the overload capacity of a synchronous motor see the reference 

 given near the end of Art. 47. 



A synchronous motor is sometimes operated at no load, and at 

 such a value of the field current that the machine draws reactive 

 leading kilovolt-amperes from the line, thus improving the 

 power-factor of the system. In such a case the machine is called a 

 synchronous condenser, or better, a phase adjuster. The diagram in 

 Fig. 41 is greatly simplified in this case because the energy com- 

 ponent of the current can be neglected, as well as the drop ir, 

 and the e.m.f. E t . We then have t=i 2 = id, and E n =E=e+ix. 

 The direct armature reaction is determined from eq. (79) in which 

 ^=90. When the motor is underexcited and draws a lagging 

 current from the line, i is to be considered negative, or ^=270 

 degrees. The same simplified diagram applies to a polyphase 

 rotary converter, operated from the alternating-current side, at no 

 load. 



Prob. 14. It is required to calculate the field current and per cent 

 voltage regulation of a 12-pole, 150 kva., 2300-volt, 60-cycle, Y-connected 

 alternator, at a power factor of 85 per cent lagging. The machine has 

 two slots per pole per phase, and is provided with a full-pitch winding. 

 the number of turns per pole per phase being 18. The armature resist- 

 ance per phase of Y is 0.67 ohm, the reactance is 3.5 ohm. The number 

 of field turns per pole is 200. The no-load saturation curve is plotted 

 for the line voltage (not the phase voltage), and at first is a straight lino 

 such that at 1800 volts the field current is 17.4 amp. The working part 

 of the no-load saturation curve is as follows : 



Kilovolts 2.2 2.4 2.5 2.6 2.7 2.78 



Field current, amp 22 25 27 30 34 40 



Ans. 31 amp.; 14.3 per cent. 



Prob. 16. Show that in the foregoing machine the short-circuit 

 current is equal to about two and a half times the rated < unvnt. at the 

 fii-M excitation which gives the rated voltage at no-load. Hint: The 

 short-circuit curve is a straight line so that one can first calculate the 

 field . in K n( for any assumed value of the armature current and -0. 



Prob. 16. From the results of the calculations of the piv.-eding 

 prol.leni -how that the cross-magi tTect ami the ohinic drop are 



negligil>l<> under short-circuit, in the machine under consideration. 



