Til/-: XY\('HK(f\(>rs MOTOR 



325 



Assume in Fig. 298 that a certain system takes 7 amperes at 



a voltage V and that the current / lags V by 0i degrees. It is 



d to raise the power-factor of the system to unity by means 



of a synchronous motor, while at the same time the motor is to 



supply mechanical power requiring V I\ watts from the line. 



The synchronous motor must first take a quadrature leading 

 current // in order to counteract the lagging quadrature cur- 

 rent 7 2 of the load. 



// = / 2 = / sin 0i 



In addition, the synchronous motor must take an energy 

 current // to supply its losses and 

 also the power required by its load. 

 The total synchronous motor 

 current 



/. = V(/i') 2 + (/ 2 ') 2 (74) 

 and tin- poucr-factor of the syn- 

 chronous motor 



j. i 

 cos 6 S = T - 



(75) 





Example. A certain machine shop 

 takes 200 kw., at 0.6 power-factor, from Fio. 298. Raising power-factor 

 a 600- volt, three-phase, 60-cycle system, to unity by means of a loaded 



:-G the power-factor to synchronous motor. 



0.0 \>y means of a synchronous motor, which at the same time is to drive 

 a direct-current shunt generator requiring that the synchronous motor 

 <> kw. from the line. What should l>e the rating of the synchronous 

 motor in volts and amperes. 



The rector diagram is >ho\vn in Kig. 299. Assume that the system is 

 Y-connected. The problem will he worked for one phase only. 



( \( M ) 



QlUge to neutral, V - -p = Jiir, volt,. 

 v 3 



The current per phase, I 



200,000 



= 321 amp. 



V't X 600 X 0.60 



HTK.V current. I, 1 cos = I X 0.0 = 192.6 limp. 

 The (jiiadrature current, I.. I >m = I X 0.8 - 257 amp. 

 At 0.9 power-fa. -for, tf - 25.8 

 }\)-2.*\ 



Cos 





e component I,' - I' sin fr - 214 X 0.436 -93.3 amp., 



:i<! Of 25.8 



