CHAPTER V. 



Design of a Three-Phase Current Motor for 2OO 

 Horse-Power. 



THE application to practice of the theory expounded in the pre- 

 ceding chapters will best be illustrated by a concrete case. To 

 this end I propose to calculate a three-phase current motor for 

 an output of 200 horse-power at 440 r. p. m. for a frequency of 60 ~. 

 Voltage between the lines 2000. I have chosen a rather extraordinary 

 case, the speed of 440 r. p. m. being comparatively low, while the 

 frequency is unfavorably high, in order to show that it is yet pos- 

 sible to build a satisfactory motor for such conditions. 



65. The following are further conditions for the design : 

 (i.) Normal output, 200 horse-power. 



(2.) Maximum torque, 400 synchronous horse-power. 



(3.) The motor must be able to start with the maximum torque. 



66. The torque of a motor is measured in mkg. or foot-pounds. The 

 product of the number of mkg. into the angular velocity of the rotor, 

 divided by 76, yields the number of horse-power available at the shaft 

 of the motor. The unit of horse-power used in America and 

 England is slightly greater than the unit used on the Continent 

 of Europe. To get "European horse-power," we should have to 

 divide by 75. As, in a great many cases, we do not care much about 

 the absolute value of the torque, we speak of torque corresponding to 

 a certain number of horse-power at a certain speed. The most nat- 

 ural speed is offered us by the speed at synchronism. If we thus 

 speak of a torque of 400 synchronous horse-power, this torque may 

 be developed at any speed, but the product of it into the angular 

 velocity of the rotor at synchronism is proportional to 400 horse- 

 power. This very convenient mode of expressing torque in "syn- 

 chronous horse-power" has been introduced by Mr. Steinmetz; at 



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