8-B] CIRCLE DIAGRAM. 279 



Line Voltage. Line Current. Watts. Apparent Watts. 



E h Wo /oV3 



185 5.9 350 I 8QO 



Power factor = W -f- / V3 = o. 185 = cos 792o'. 



No-load copper loss per phase = RJ 2 = 0.255 XS : 9 2 = 8 - 8 7' 

 watts. 



No-load copper loss total = 3# 1 / 2 = 3 X 0.255 X S& = 2 ^ 

 watts. 



Iron loss, friction and windage 350 26.6 = 323.4 watts. 



4. Locked Readings. With the rotor locked in position, by 

 any convenient means, the motor is a stationary transformer on 

 short circuit. The locked or short-circuit readings, taken at re- 

 duced voltage, are 



E. I. W. Apparent Watts. Power Factor. 



54 16.5 581 1544 0.375 = cos 68 



These readings were taken at about full load current and % 

 normal voltage. At normal voltage the current would have been 

 excessive, so the desired readings for normal voltage are found 

 by proportion to be as follows : 



CALCULATED READINGS FOR FULL VOLTAGE. 



E. /s- #s. Apparent Watts. Power Factor. 



185 56.6 6810 18150 0.375 = cos 68 



These calculations are made by taking current as proportional to 

 voltage, and power as proportional to voltage squared; power 

 factor remains the same. These relations are not strictly true 

 (see 323) and it is better, therefore, to take readings for several 

 currents and to determine the results by averages or by means of 

 curves which are extrapolated for full voltage. These results 

 depend upon the temperature at which the measurements are 

 made ; excessive current should, therefore, be avoided. 



by Charters and Hillebrand, Reduction in Capacity of Polyphase Motors 

 due to Unbalancing in Voltage, A. I. E. E., Vol. XXVIII., p. 559. 



