194 ELECTRICAL MACHINERY 



Example of Efficiency Prediction. The readings ob- 

 tained from such a test upon a 100-h.p. motor would give 

 results about as follows: 



Rated voltage = 230 volts; 



Rated current =350 amperes; 



Shunt field resistance (hot) =75.2 ohms; 

 Armature resistance (hot) =.0112 ohm; 



Armature current (running light) = 15.1 amperes; 

 Area of brush contacts =70 sq.cm.; 



Current density full load =5 amperes per sq.cm.; 



Drop at the brush contacts (full load) 



= .8+02X5) = 1.80 volts; 

 Drop at the brush contacts (half load) 



= .8+(.2X2.5) = 1.30 volts, etc,; 

 The armature PR loss at no load 



= 15.1 2 X. 0112 = 2.5 watts; 



Stray power (no load) 



= (230X15.1)- |2.5+15.1/.8+(.2X 1 l ^ 1 )) j =3465 watts; 

 Shunt field loss f = 1060 watts. 



From the data the curves of Fig. 115 were plotted. 

 Then the total loss curve was constructed and from this 

 the efficiency was readily computed. The shape of the 

 efficiency curve is about the same for any electric motor 

 or generator. The efficiency is low at light loads, reaches 

 a fair value at half-load and, from this point up to one and 

 one-quarter load, is practically constant. The full-load 

 efficiency of large machines may be as high as 94%, while 

 for small machines of a few horsepower it is nearer 80%. 



