POWER LOSSES IN GENERATORS AND MOTORS. 133 



age, it is easy to calculate the true efficiency of the dynamo for 

 a specified output of power as a generator at the given speed and 

 voltage, the resistances of the various windings of the dynamo 

 being known. This calculation is based upon the equation 



power output 

 efficiency = 



power output -f losses 



It is much more instructive to discuss this matter of efficiency 

 calculation by means of several typical numerical examples, than 

 by deriving a general formula. 



Example i. The shunt generator. Given a shunt generator 

 which delivers 50 amperes of current at 1 10 volts between its 

 terminals. It is required to calculate the efficiency of the gen- 

 erator under these conditions having given the following data : 



R s = 44 ohms (hot) including the portion of the field rheostat 

 which must be in circuit to bring the voltage between 

 the terminals of the machine to the specified value. 



R a = o. 1 4 ohm (hot). 



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



Solution : 



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



iE\ 2 



(b) Field loss = R s x ( ~] =275 watts 



/ \ 2 



(c) Armature loss= A* ( 50 + ^f ) =386 watts 



(d) Efficiency = - - = 0.801 



5,500 + 275 + 386 + 700 



Example 2. The compound generator (long-shunt). Given a 

 long-shunt compound generator which delivers 50 amperes of 

 current at 1 10 volts between its terminals. It is required to cal- 

 culate the efficiency of the generator under these conditions hav- 

 ing given the following data : 



