

DETAILED STUDY OF OPERATION WITH NORMAL LOAD 71 



tLet us suppose, now, that each group of motors shall furnish 50 

 v. The active current for the induction-motors can be estimated 

 at 25 amperes, plus. 8 amperes for losses, or in all 33 amperes; and 

 the active current for the synchronous motors can be estimated at 

 25 amperes, plus the losses, which are at most 10 amperes, or, in all, 

 35 amperes. 



The total active current is therefore 33 + 35=68 amperes, while 

 the reactive current of the induction-motors remains, as before, 42 

 amperes. The resultant current is 



v / (42) 2 +(68) 2 =g2.5 amperes, 



and it could not be less than 68 amperes. 



Suppose it be required to bring up the power-factor to 0.95, i.e., 

 to bring the current down to 72 amperes. It will be sufficient to 



68 

 reduce the reactive current to =23 amperes by producing a current 



o 

 out of phase which is equal to 



/= 42 23 = 19 amperes. 



The current of the synchronous motors will then become equal 

 to about 



V / (i9) 2 +(35) 2 =4o amperes 



instead of 35 (i.e., it would be slightly increased). We will have 

 tan r=$$ =0.543 and the E.M.F. will be equal to 



3= 2000 J I + (^^V + ^-(19X0.4771 -35X0.8788) 

 ^ \ 2OOO / . 2OOO 



= 1728 volts. 



instead of 1486 volts, which the motor should develop, when the phase- 

 angle f equals zero. 



It is seen that, by overexciting to a maximum of 30 per cent, it is 

 possible to compensate for the full-load reactions, without overloading 

 the synchronous motors. To obtain the same result by means of a 

 motor running without load it would be necessary to use an apparatus 

 having a rating of about 40 k.w.; which would not be very practical. 



