2/2 INDUCTION MOTORS. [Exp. 



44. The primary copper loss is computed* from the measure- 

 ments of current and resistance. 



45. Resistance Measurements on Generator. Immediately 

 after the run, measure the resistance of the generator armature, 

 including brushes and leads up to the voltmeter terminals. 



46. Calculation of Motor Output. The output of the motor is 

 equal to the generator output El, plus the generator and belt losses. 

 The losses are found as follows: 



The armature copper losses RI 2 of the generator are calculated 

 for the particular armature current, the resistance being deter- 

 mined as in 45. 



The field copper loss does not enter into the computations when 

 the generator is separately excited. (When the generator is self- 

 excited, the loss in the field circuit and in the field rheostat is 

 determined by the product of field current and terminal voltage.) 



The belt loss and the rotation losses of the generator are deter- 

 mined as in 41, or by a separate test made as in Exp. 2-B (see 

 particularly 21, 24, Exp. 2-B). The method of 41 is, how- 



* (44a). In a 3-phase motor, for example, in which the measured line 

 current is 10 amperes and the measured resistance R between terminals 

 is i ohm, the copper loss is computed from the equivalent single-phase 

 current and resistance as follows : 



Equivalent single-phase resistance R' = .$ ohm. 

 Equivalent single-phase current /' = 10 V~3 = 17.3 amperes. 

 Total copper loss = ^7 /2 = .5 X 300= 150 watts. 



The computations may also be made by assuming the motor to be either 

 star or delta connected, as follows. 



Assuming a star connection, the resistance per phase is .5 ohm and the 

 current per phase is 10 amperes. The RI* loss per phase is accordingly. 

 50 watts; the total loss is 150 watts as before. 



Assuming a delta connection, the current per phase is 10 *- V* 3 = 5-77 

 amperes; the resistance per phase is 1.5 ohms. The /?/* loss per phase is 

 50 watts and the total loss is 150 watts as above. This numerical example 

 shows that the same result is obtained by the three methods of computa- 

 tion; it is unnecessary to know whether the primary is delta or star 

 connected. 



