326 THE ROTARY CONVERTER. 



of points to the voltage given on the observed 

 curve. The dotted curve is seen to become almost 

 horizontal after an initial drop from the true no-load 

 value. The slight inclination of the straight part of the 

 curve must be ascribed to armature magnetic reactions, or 

 the screening effect of eddy currents in the pole faces. 

 An example of the calculation of one point on the dotted 

 curve will make the method of obtaining it clear. 



Take the point corresponding to 12 amps. A.C. From 

 Fig. 162 it is seen that the direct current supplied is 

 10'7 amps. 



Of this 2'5 amps, are spent in driving the machine, 

 and 8 '2 are converted into alternating current. We there- 

 fore have for the total ohmic drop (referring to the table 

 above), since the armature resistance is '145 ohms, due to 

 driving current 2'5 x -145 = '36 volts, 



converted current 8'2 x -145 x 1-175 = 1'4 



Total drop . . . . 1'76 



It is to be noted that the machine for which the 

 curves are drawn was an experimental machine of small 

 size. The losses due to resistance of the armature and 

 friction of the machine appear unduly high in conse- 

 quence. In a machine of commercial size when working 

 under normal load, the losses due to the no-load currents 

 would be entirely negligible, and the total loss in volts 

 would be obtained directly by employing the formula on 

 p. 325. 



From Fig. 162, which shows the ratio of current 

 transformation, it is seen that the ratio is a constant one, 

 since the curve is a straight line. It cuts the vertical axis 

 at 2'5 amps., showing that this was the current required 

 to drive the converter when running light and supplied at 

 the normal voltage 011 the direct-current side. The power 

 taken under these conditions ( = 2*5 x 100 = 250 watts in 

 the present case) represents the power required to over- 

 come friction, windage and no-load iron losses. The iron 

 losses will increase somewhat under load, but approxi- 

 mately the no-load losses + the copper losses, which can 

 be calculated for any load, will give the total losses at any 

 load. The approximate efficiency of the converter when 

 driven from the direct-current side could therefore be 

 calculated from the no-load observation. 



