SURGING OF SYNCHRONOUS MOTORS
hence, for small values of s:

291

dM
dt'

dt

vSince:

(6)

and from (5):

it is:

d8

s = de'

ds _ d2d

de ~~ cW1

dM

(7)

Since, as discussed, the change of momentum equals the dif-
ference between produced and consumed power, the excess of
power being converted into momentum, it' is:

p-n-f. (8,

and, substituting (4) and (7) into (8) and rearranging:

II = 0- (9)

Assuming 5 as a small angle, that is, considering only small
oscillations, it is:

. 6 §
sin g = 2*

sin \a — ft — s = sin (a — /5);
hence, substituted in (18):

— 8 sin (a — /3) + 4 irfMo -^ = 0,
and, substituting:-

a =

eep sin (a

(10)
(ID
(12)