132 THE DTRECT-rrRRTWT MOTOR < 'II. VI 



intensity of magnetisation is 1 1,000 lines |MT s<|iiaiv centi- 

 metre. Find the torque required to turn the armature 

 when the magnets are excited. Suppose that from our 

 knowledge of the quality of the iron used, we take h to 

 be 9,000. The iron in the armature of a ten pole machine 

 makes 5 complete cycles per revolution, giving *=.">. 

 Hence the required torque is 3,940 inch-pounds, and this 

 must be continually exerted in order to turn the armature 

 against the retarding torque due to hysteresis. The 

 dynamo is an 800 k.w. Westinghouse Railway Generator. 



When a motor is placed on a circuit without any 

 external load, and allowed to run by itself, the load on 

 the motor is the sum of all the retarding torques that 

 we have been considering. Hence, if we observe the 

 current and know the value of 37, we can find the torque 

 losses. 



The point of intersection of the curve of useful torque 

 and the axis of current gives us the value of the current 

 the motor will take when running with no external load. 

 Thus, in Fig. 30, this current is given by the intercept bh. 



Example 33. A motor when placed on a line of 

 100 volts tension runs at 1,240 r.p.m. and takes a 

 current of 4 amperes. If the resistance is O'l ohm, find 

 the torque losses. From the usual equation for the 

 speed, we find the value of M to be 4'82, and since the 

 current is 4 amperes, the torque is 27 inch-pounds. As 

 the external load is nothing, the whole of this torque 

 is expended in friction, hysteresis and eddy current loss. 



The ratio of the useful mechanical watt output to the 

 total electrical watt input is the total efficiency of the 

 motor. If this ratio is plotted on the axis of the current, 

 it will start from the point h, in Fig. 30 ; and rise to a 



