THE MAGNETIC FIELD IN THE POLYPHASE MOTOR. 



(8) 61.6 . D m ks = -^~ (3/ 2 *r 2 + />,*,). 



i 



If we want to have the torque in foot-pounds, the formula is 



8.5 . Dft. ibi. = -^- (3 *j * r 2 + P wat ts). 



~i 

 The torque is, therefore, proportional to the algebraic sum of the 



output of the motor plus the energy dissipated in the armature. 



26. Starting Torque. At starting P is zero. All the energy that goes 

 into the motor, is dissipated into heat ; therefore. 3 (z\ 2 r t -f- i t * r t ) =- 3* 

 must be equal to the ordinate of our circle. If TI and r 2 are known, 

 the point on the circle corresponding to this case can easily be found 

 by trying. 



A glance at our diagram shows us that we can attain at starting any 

 running torque ; but the starting current will always be equal to, and 

 never smaller than, the running current, 4 unless the motor is started 

 under conditions, as higher or lower voltage, etc., different from those 

 under load. 



27. Efficiency. The energy which flows into the motor neglecting 

 hysteresis can be taken from the diagram. If e\ is the impressed 

 e. m. i., ii the current in each phase, then we have 



(9) = 3'i*iw* 



To obtain the output we have to deduct the ohmic losses, the load 

 losses, and the friction. Hysteresis and eddy currents may be taken 

 into account, since they are practically constant at all loads, as though 

 they were ohmic losses in a coil so placed upon the primary as to have 

 no leakage at all, the energy wasted in this coil being equal to the 

 loss through hysteresis and eddy currents. We have, then, 



(10) P = 3 ^ /! cos * 3 t\ ' r t 3 i, > r t F Q 



4 Prof. Silvanus P. Thompson, "Polyphase Electric Currents," first edition, p. 

 213, says of a polyphase motor: "This motor starts under full load, taking less 

 than full-load current." This is an impossible figment. In the second, vi-ry imu-h 

 enlarged, edition of Prof. Thompson's interesting honk, this statement is changed, 

 and we read on page 255: "This motor starts under full load, taking less than 

 twice the full-load current." The motor should theoretically not take more than 

 full-load current, but just as much, and in practice this is also generally obtained, 

 as, in this case, the rotor is equipped with slide rings. 



17 



