THE INDUCTION MOTOR. 



whence follows 



j) 

 (14) ..... 6i.6D m k e = - J - . P-watts, 



~i 



in which equation Dmkg is the torque in m.kg., and p the number 

 of north or south poles. 



102. The following reasoning yields a value for ~, : 



We have for motor 7, 



9.81 . A (i - s) = 3 *' a ' r> , 

 and for motor II, 



9.81 . DnK + u,) = 3 '"' * 2 ' r \ . 



in which equations t is the angular velocity of the revolving field, 

 <>, that of the armature. 

 Hence, 



9.81 . (Di - Dn ) " 2 = 3 ^ 



\" 2 / 



or 



(IS) ..................... 3 *, * r* = Plaits ~ 



803. To illustrate : Let us assume ~i = 50, and ~ 2 = 45 ; then we 

 have 



= 0.23 P-watts 



In words, if the slip is 10 per cent, the loss of energy in the arma- 

 ture amounts to 20 per cent. 



104. It is instructive to compare with (14) the formula (7) for the 

 polyphase motor, which reads after some transformations, 



(16) ..................... 3 t\ * r z = P-watts . - 



~ 



This is in words that the slip in per cent is equal to the armature 

 loss in per cent, setting P + 3 it* r* equal to 100. 



58 



