192 ELECTRICAL MACHINERY 



the losses occurring in the machine; hence the efficiency 

 may be obtained by using the formula 



. output 



Efficiency = . ... (37) 



output + losses 



At full load the output of the motor referred to in the pre- 

 vious paragraph is equal to 746X5 = 3730 watts. 



The input must be 3730 +(242 +97 +80 +320) =4469 watts. 



qyon 



Hence the full load efficiency = = 83.6%. 



2982 

 When the output = 4 h.p., the efficiency = 



2932+627 t 



oooo 



When the output = 3 h.p., the efficiency = = 80.3% 



~~ 



When the output = 2 h.p the efficiency =7-^ =75.1% 



- 



746 

 When the output = 1 h.p., the efficiency = -, =62.1% 



/ TtO ~j- 4OO 



When the output = h.p., the efficiency = jon = 0% 



Ordinarily the loss curves are not obtained in terms of 

 horsepower output but in terms of the armature current 

 and hence the efficiency curve is obtained in terms of the 

 armature current. 



Efficiency Determination without Actually Loading the 

 Machine. To determine the efficiency of an electrical 

 machine it is not necessary to actually load it; some method 

 of determining the various losses is all that is required. 



