### Notes on the Theory of the Synchronous Motor, With Special Reference to the Phenomenon of Surging

Charles Proteus Steinmetz
1902 Transactions of the American Institute of Electrical Engineers
While an induction motor at constant impressed voltage is fully determined as regards current, power factor, efficiency, etc., by one independent variable, the load or output, in the synchronous motor two independent variables exist, load and field excitation. That is, at constant impressed voltage the current, power factor, etc., of a synchronous motor may at the same power output be varied over a wide range by varying the field excitation, that is, the counter e.m.f. or "nominal induced
more » ... minal induced e.m.f." Hence the synchronous motor may be utilized to fulfill two independent functions, to carry a certain load and to produce a certain wattless current, lagging by under-excitation, leading by over-excitation. Synchronous motors are, therefore, to a considerable extent used to control the phase relation and thereby the voltage, in addition to producing mechanical power. The same applies to synchronous converters. With given impressed e.m.f., field excitation or counter e.m.f. corresponding thereto, and load, determine by the usual theory all the quantities of the synchronous motor, as current, power factor, etc. Thus, if in diagram Fig. 1 , 0 Ee = e.m.f. consumed by the counter e.m.f. of the synchronous motor as corresponding to its field excitation, and if Po = output of motor (exclusive of friction and core loss and, if the exciter is driven by the motor, power consumed by the exciter), i,1-P0/e 781 782 STENNMETZ: SYNCHRONOUS MOTOR [June 1I energy component of current, represented by 0 1" and the current vector therefore must terminate on a line i perpendicular to O I. If, then, r = resistance and x = reactance of the circuit between counter e.m.f. e and impressed e.m.f. e0, Q Er i r = e.m.f. consumed by resistance, 0 Ei i,x = e.m.f. consumed by reactance of the energy current i, hence 0 E'l, e.m.f. consumed by impedance of the energy current i" and the impedance voltage of the total current lies on the perpendicular e on O El. Producing 0 E, OE, and drawing an arc with the impressed e.m.f. eo as radius and E, as center, the point of intersection with e' gives the impedance voltage OF1, and corresponding thereto the current 0 I-i; and completing the parallelogram 0 E E, E' gives the impressed e.m.f. 0 EoF