THE INDUCTION MOTOR. 415 



the no-load current. Next, select a suitable ampere 

 scale for the diagram, say one centimetre corresponds to 

 20 amperes. Make, therefore, the piece K - 2 cms. 

 = C , and make the piece B l =26-6 cms. = <7 S . 

 Now draw the semi-circle A B F, having its centre along 

 the line O A (viz., at point p), and passing through the 

 points B l and K already found. Through point K 

 draw the line a 6 parallel to O A. -^ 



To get the circle of the output, the output of the 

 motor is zero for the point B 1 , as the motor is then at 

 rest. Therefore join this point to point A, and from 

 the latter draw the line A f, so that the angle B 1 A f is 

 90 ; the point / so found is the centre of the " output " 

 semi-circle which can now be drawn at such a radius that 

 it passes through the points A and F. The line B 1 A 

 is thus a tangent to the circle A N F, so that the value 

 of the output for the point B 1 corresponding to the 

 stator current of 532 amperes, is zero. 



To find the circle for the torque, scale off the values 

 B 1 A and O A from the diagram, which will be found to 

 be 6-4 and 27-5 cms. respectively; the latter corre- 

 sponds to a pressure per phase of 202 volts. Hence 



Value of vector B 1 A = J jj, x 202 = 47 volts. 



Z i ' o 



The copper drop per phase of the stator for the 

 current C 8 == 532 amperes is (532 x 0-032) == 17 volts. 

 Hence, to get the desired point, H l on the torque circle, 

 we mark off along B 1 A from the point B 1 a piece equal 



17 x 6-4 

 to TT~~~ units of length, thus equal to 2-31 cms. 



Hence the torque circle is fixed by finding the point H l ; 

 from a suitable centre along p f (viz. f 1 ) draw it 

 through points A, H L , and F. 



Finally, to get the slip line, drop a perpendicular 

 from the point B 1 on the radius A f 1 , thus getting the 

 line of B 1 T, which turns out to be 6 25 cms. in length. 

 This is equal to a slip of 100 per cent, as the motor is 

 not running ; at the load corresponding to the stator 

 current B, torque H Zr, and output N P, the slip is 

 equal to the piece V T cut off on the slip line B 1 T, 

 which scales 0-25 cm. Consequently the full load slip 

 is 4 per cent. ; at the maximum load the motor will 

 carry (156 b.h.p.) the slip is 16 per cent. 



