POWER LOSSES IN GENERATORS AND MOTORS. 135 



terminals. It is required to calculate its efficiency under these 

 conditions, having given the following data : 



R c = o. 1 2 ohm (hot) 

 R a = o. 1 5 ohm (hot) 



Stray power loss at given speed and voltage = 700 watts. 

 Solution : 



(a) Power output =50 amperes x 1 10 volts = 5,500 watts 



(b) Series field loss = RJ* = 300 watts 



(c) Armature loss =RJ*= 375 watts 



(^Efficiency = 



66. Efficiency of conversion and electrical efficiency of a genera- 

 tor. The total mechanical power expended in driving the arma- 

 ture of a generator is used : (a) to supply the stray power loss S, 

 that is, to overcome the opposition to the rotation of the arma- 

 ture due to friction, windage, eddy currents, and hysteresis ; and 

 (fi) to overcome the opposition to the rotation of the armature 

 due to the action of the field magnet on the armature conductors 

 on account of the current flowing in them. The portion (&) of 

 the total power is, according to Lenz's law, equal to the total 

 electrical power E a I aJ developed in the armature. The ratio of 

 the total electrical power, ' EJ aJ developed in the armature to 

 the total mechanical power supplied to the armature, is called the 

 efficiency of conversion of the generator. From the above discus- 

 sion the total mechanical power delivered to the armature is evi- 

 dently (EJ a + S), so that 



El 



Efficiency of conversion = _. .. a " ~ (3 1) 



A/a ' ^ 



A portion, A, of the total electrical power developed in the 

 armature of a generator is used to overcome armature resistance, 

 a portion, F t is expended in field excitation, and the remainder 

 EJ a A F is delivered as useful power to the external re- 



