86 TIIK I>li;K('T-<Tl;l;KNT .M<>To|; CII. IV 



testing of a motor which has to comply with specified 

 conditions. 



Example 27. A motor has to be tested while run- 

 ning at 900 r.p.m. on a line of 125 volts tension, the 

 torque on the motor shaft being 3,150 inch-pounds and 

 the resistance of the motor 0-015 ohm. A second dynamo 

 is provided similar in all respects and having the same 

 resistance as the motor to be tested. Equation 1 8 gives 

 us the value of the induction factor of a motor running 

 on a line of given tension, with a specified torque, and at 

 a given speed. From this equation we find the value of 

 the induction factor of the motor to be 8 '00 ; it will then 

 be taking 278 amperes. Using Equation 43 we find the 

 induction factor of the generator to be 8'61(J. The torque 

 required to make up the heat loss will be 218 inch- 

 pounds. If the dynamometer indicates 600 inch-pounds, 

 there is a torque loss in the motor of 191 inch-pounds. 



Returning now to the original arrangement where 

 two dynamos are connected in parallel on a line of constant 

 tension. We found that when the resultant torque on 

 the shaft was nothing, there was a current passing from 

 the line representing the energy required to make up the 

 heat losses in the two machines. The current then 

 passing in the motor is greater than that in the generator 

 by the amount of the current from the line. If there is 

 no other loss than that due to heat, the currait from the 

 line will be given by Equation 32. If, however, there 

 are losses due to friction, hysteresis, &c., the line current 

 will be greater than this, and the difference will represent 

 these losses. We have here a modification of Hop- 

 kinson's method of testing, namely one in which the 

 losses are measured electrically. 



