CH. VI EFFICIENCY 143 



car equipped with motors having the higher induction 

 curve will run the slower of the two. 



Example 39. The weight of the car of a passenger 

 lift was balanced until the motor took the same current 

 ascending and descending. The current at full speed was 

 then 20 amperes, the tension at the terminals of the motor 

 120 volts, and the resistance O'l ohm. This current 

 represents the torque loss at no load. A load of 1,600 

 pounds was now placed in the car, and lifted at the rate 

 of 126 feet per minute, the motor then taking 73 amperes. 

 The mechanical output is 4,560 watts, and the corresponding 

 electrical input 8,760 watts, giving a total efficiency of 

 52 per cent. If we calculate the efficiency by Equation 

 57, taking (7=20, we get an efficiency of 68 per cent., 

 showing that the torque loss increases with the load. 

 By inserting in Equation 57 the efficiency as found 

 by experiment, we see that c f is 32 amperes, showing that 

 the torque loss increases by rather more than 50 per cent, 

 from no load to full load. In this case the velocity ratio 

 was 70, the induction factor 4 '6, and the diameter of the 

 rope drum 24 inches. If the induction factor and the 

 diameter of the rope drum are both doubled, we get the 

 same car speed on the same tension, but with increased 

 efficiency. If at the same time we reduce the resistance, 

 say to 0'05 ohm, we increase the efficiency still more. 

 The value of e f is reduced to 16 amperes, and the 

 efficiency increased to 75 per cent. 



The following method for determining the 

 efficiency of series- wound motors at different loads 

 will be found useful. 



Two similar motors are geared together, the pinions of 

 each meshing into one and the same intermediate gear 



