CHAPTER XVIII 
SURGING OF SYNCHRONOUS MOTORS 

166. In the theory of the synchronous motor the assumption 
is made that the mechanical output of the motor equals the power 
developed by it. This is the case only if the motor runs at 
constant speed. If, however, it accelerates, the power input is 
greater; if it decelerates, less than the power output, by the power 
stored in and returned by the momentum. Obviously, the 
motor can neither constantly accelerate nor decelerate, without 
breaking out of synchronism. 

If, for instance, at a certain moment the power prod wed by 
the motor exceeds the mechanical load (as in the moment of 
throwing off a part of the load), the excess power is consumed by 
the momentum as acceleration, causing an increase of speed. 
The result thereof is that the phase of the counter e.m.f., c, 
is not constant, but its vector, e, moves backward to earlier time, 
or counter-clockwise, at a rate depending upon the momentum. 
Thereby the current changes and the power developed changes 
and decreases. As soon as the power produced equals the load, 
the acceleration ceases, but the vector, c, still being in motion, 
due to the increased speed, further reduces the power, causing 
a retardation and thereby a decrease of speed, at a rate depend- 
ing upon the mechanical momentum. In this manner a periodic 
variation of the phase relation between e and to, and correspond- 
ing variation of speed and current occurs, of an amplitude and 
period depending upon the circuit conditions and the mechanical 
momentum. 

If the amplitude of this pulsation has a positive decrement, 
that is, is decreasing, the motor assumes after a while a constant 
position of e regarding ea, that is, its speed becomes uniform. 
If, however, the decrement of Hie pulsation is negative, an 
infinitely small pulsation will continuously increase in amplitude, 
until the motor is thrown out of step, or the decrement becomes 
zero, by the power consumed by forces opposing the pulsation, 
as anti-surging devices, or by the periodic pulsation of the syn- 
chronous reactance, etc. If the decrement is zero, a pulsation 
288 


SURGING OF SYNCHRONOUS MOTORS 289 

started once will continue indefinitely at constant amplitude. 
This phenomenon, a surging by what may be called electro- 
mechanical resonance, must be taken into consideration in a 
complete theory of the synchronous motor. 
167. Let: 

E0 = e0 = impressed e.m.f. assumed as zero vector. 
E = e (cos P — j sin P) = e.m.f. consumed by counter e.m.f. 
of motor, where: 

P = phase angle between E0 and E. 
Let: 

Z = r + jx, 

and z = Vr2 + x2 

= impedance of circuit between 
Eo and E, and 

x 
tan a = — 
r 

The current in the system is: 

e0 — E eo — e cos P + je sin P 


/o = 


r + jx 
= - {[e0 cos a — e cos (a + P)] 


— j [e0 sin a — e sin (a + 0)] ) (1) 
The power developed by the synchronous motor is: 

Po = [EI]1 = - {[cos p [e0 cos a - e cos (a + 0)] 

z 

+ sin 0 [e0 sin a — e sin (a + 0)] J 

= {[e0 cos (a — 0) — e cos a]). (2) 

If, now, a pulsation of the synchronous motor occurs, resulting 
in a change of the phase relation, 0, between the counter e.m.f., e, 
and the impressed e.m.f., e0 (the latter being of constant fre- 
quency, thus constant phase), by an angle, 5, where 8 is a periodic 
function of time, of a frequency very low compared with the 
impressed frequency, then the phase angle of the counter e.rn.f., 
e, is P + 6; and the counter e.m.f. is: 

E = e {cos (0 + 6) - j sin (p + 6)1, 

19 


290 ELECTRICAL APPARATUS 

hence the current: 

/ = - {[e0 cos a — e cos (a + 0 + 5)] 

z 

— j [e0 sin a — e sin (a + 0 + 6)]\ 

= h + ysin* jsin(a + p+ *) + jcos(a + 0 + |) [ (3) 

the power: 

i 

P = {e0 cos (a — 0 — 5) — e cos a} 

= Po + -— sin ^ sm^a - 0 - 2 J • (4) 

Let now: 

t»o = mean velocity (linear, at radius of gyration) of syn- 
chronous machine; 

a = slip, or decrease of velocity, as fraction of t'0, where s is 
a (periodic) function of time; hence 

v = Vq (1 — s) = actual velocity, at time, t. 

During the time element, dt, the position of the synchronous 
motor armature regarding the impressed e.m.f., e0, and thereby 
the phase angle, 0 + 6, of e, changes by: 

dd = 2 Tcfsdt 

= sd0, (5) 

where: 

0 = 2 icft, 

and 

/ = frequency of impressed e.m.f., e0. 
Let: 

m = mass of revolving machine elements, and 

M0 — )i im-'o2 = mean mechanical momentum, reduced to 

joules or watt-seconds; then the momentum at time, t, and 

velocity v = v0 (1 — s) is: 

AT = y2vivj(\ - s)2, 

and the change of momentum during the time element, eft, is: 

dM , .ds. 


svmsixg or srxcHROsors motors 2*1 


hence. 

for srxihiL 

TATaes Oil 

*: 

= — 

.d* 

™rdi 

"Mld* 

at 
ds 
d* 
dt 

Since: 

4« 
di 

= 2* 

and from 

5 : 

t 

di 
d« 

<U 

<r-h 

d9 

dr- 

it is: 

d\I 

di 

= — 

4„.«.£ 

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 - Po = d£ • l» 

and. substituting <'4> and (7) into (8) and rearranging: 

C*-° sin * sin (a - fi - *) + 2 t/AT 0 ™ = «• l»> 

'Assuming 5 as a small angle, that is, considering only small 

oscillations, it is: 

. 6 6 

Sln2 = 2 


sin [a - 0 - ^ J = sin (« - £) ; 


hence, substituted in (18): 


^ 5 sin (« - 0) + 4 ir/Af 0 jj]» - 0, (10) 


and, substituting: 


ce0«in(a-/8) m. 

4 rfzMo 


it is: 


(14) 


292 ELECTRICAL APPARATUS 

This differential equation is integrated by: 

5 = At'9, (13) 

which, substituted in (12) gives: 

aAece + ACU™ = 0, 
a + C2 = 0, 

C = ± V- a. 
. 168. 1. If a <0, it is: 

5 = A*+me + A*-m\ 
where: 

/ / ee0 sin (0 — a) 

\ 4 tt/zA/o 

Since in this case, e*m9 is continually increasing, the syn- 
chronous motor is unstable. That is, without oscillation, the 
synchronous motor drops out of step, if 0 > a. 

2. If a > 0, it is, denoting: 

, ^/- , /ee0sin (a - 0) 

\ 4 TT/^Af o 

or, substituting for €+;n* and €+""4* the trigonometric functions: 

6 = (Ai + Ao) cos n0 + j (Ax — *42) sin n$t 
or, 

5 = B cos (n0 + 7). (15) 

That is, the synchronous motor is in stable equilibrium, when 
oscillating with a constant amplitude B, depending upon the 
initial conditions of oscillation, and a period, which for small 
oscillations gives the frequency of oscillation: 

f „f _ //ee0 sin (a - 0) 

As instance, let: 

<?o = 2200 volts. Z = 1 + 4 j ohms, or, z = 4.12; a = 76°. 

And let the machine, a 16-polar, 60-cycle, 400-kw., revolving- 
field, synchronous motor, have the radius of gyration of 20 in., 
a weight of the revolving part of 6000 lb. 

The momentum then is Af„ = 850,000 joules. 

Deriving the angles, 0, corresponding to given values of output. 
P, and excitation, r, from the polar diagram, or from the symbolic 


SURGING OF SYNCHRONOUS MOTORS 293 

representation, and substituting in (16), gives the frequency of 
oscillation : 

P = 0: 

e = 1600 volts; 0 = - 2°;/0 = 2.17 cycles, 

or 130 periods per minute. 


2180 volts 

+ 3° 

2.50 cycles, 

or 150 periods per minute. 

2800 volts 

+ 5° 

2.85 cycles, 

or 169 periods per minute. 

P = 400 kw. 

e = 1600 volts; 0 = 33°; /0 = 1.90 cycles, 

or 114 periods per minute. 
2180 volts -21° 2.31 cycles, 

or 139 periods per minute. 
2800 volts 22° 2.61 cycles, 

or 154 periods per minute. 

As seen, the frequency of oscillation does not vary much with 
the load and with the excitation. It slightly decreases with 
increase of load, and it increases with increase of excitation. 

In this instance, only the momentum of the motor has been 
considered, as would be the case for instance in a synchronous 
converter. 

In a direct-connected motor-generator set, assuming the 
momentum of the direct-current-generator armature equal to 
60 per cent, of the momentum of the synchronous motor, the 
total momentum is M0 = 1,360,000 joules, hence, at no-load: 

P = 0, 
e = 1600 volts ;/0 = 1.72 cycles, or 103 periods per minute. 

1.98 cycles, or 119 periods per minute. 
1.23 cycles, or 134 periods per minute. 

169. In the preceding discussion of the surging of synchronous 
machines, the assumption has been made that the mechanical 
power consumed by the load is constant, and that no damping 
or anti-surging devices were used. 

The mechanical power consumed by the load varies, however, 
more or less with the speed, approximately proportional to the 
speed if the motor directly drives mechanical apparatus, as 
pumps, etc., and at a higher power of the speed if driving direct- 
current generators, or as synchronous converter, especially 


294 ELECTRICAL APPARATUS 

when in parallel with other direct-current generators. Assum- 
ing, then, in the general case the mechanical power consumed by 
the load to vary, within the narrow range of speed variation con- 
sidered during the oscillation, at the pth power of the speed, 
in the preceding equation instead of Po is to be substituted, 
Po(l - s)p = P0(l - ps). 

If anti-surging devices are used, and even without these in 
machines in which eddy currents can be produced by the oscilla- 
tion of slip, in solid field poles, etc., a torque is produced more 
or less proportional to the deviation of speed from synchronism. 
This power assumes the form, Pi = c2s, where c is a function of 
the conductivity of the eddy-current circuit and the intensity 
of the magnetic field of the machine, c2 is the power which 
would be required to drive the magnetic field of the motor 
through the circuits of the anti-surging device at full frequency, 
if the same relative proportions could be retained at full fre- 
quency as at the frequency of slip, s. That is, Pi is the power 
produced by the motor as induction machine at slip «. In- 
stead of P, the power generated by the motor, in the preced- 
ing equations the value, P + Pi, has to be substituted, then: 

The equation (8) assumes the form : 

P + Pi-PoU-jmO = rfJ^ 

or: 

(P - Po) - (P. + pPos) = *% • (17) 

or, substituting (7) and (4) : 

2ee2° sin | sin [« - 0 - *] + (c« + pPo) * + 4 ,/Af. ~ = 0; 

(18) 
and, for small values of 8 : 

4 vfzMo 
b , «! + PpJL. (2Q) 

Of these two terms b represents the consumption, a the oscilla- 
tion of energy by the pulsation of phase angle, p. b and a thus 


SURGING OF SYNCHRONOUS MOTORS 295 

have a similar relation as resistance and reactance in alternating- 
current circuits, or in the discharge of condensers, a is the 
same term as in paragraph 167. 

Differential equation (19) is integrated by: 

5 = Atc', (21) 

which, substituted in (19), gives: 

aAtc* + 2 bCAf + C2Aec* - 0, 

a + 2 bC + C2 = 0, 

which equation has the two roots: 

Ci - -6 + Vb*-a, 

C, = -6 - y/b2 - a. (22) 

1. If a < 0, or negative, that is 0 > a, C\ is positive and C% 
negative, and the term with C\ is continuously increasing, that 
is, the synchronous motor is unstable, and, without oscillation, 
drifts out of step. 

2. If 0 < a < b2, or a positive, and b2 larger than a (that is, 
the energy-consuming term very large), C\ and C% are both 
negative, and, by substituting, + \/b2 — a = g} it is: 

Ci= - (6-flf), C, = - (6 + g); 
hence: 

5 = Al€ " <»-*>• + A2€ " (» + •)•• (23) 

That is, the motor steadies down to its mean position logarith- 
mically, or without any oscillation. 

b2 > a, 
hence : 

(c2 + pPo)2 eeo sin (a - 0) 
16wfMo > 2 (M) 

is the condition under which no oscillation can occur. 

As seen, the left side of (24) contains only mechanical, the 
right side only electrical terms. 

3. a > b2. 

In this case, y/b2 — a is imaginary, and, substituting: 

g = y/a'-b*, 
it is: 

t\=-b+jg, 


296 ELECTRICAL APPARATUS 

hence : 

and, substituting the trigonometric for the exponential functions, 
gives ultimately: 

6 = Bt-b°cos(ge + y). (25) 

That is, the motor steadies down with an oscillation of period: 

/o-flf 


■v 


_ fee0 sin (a - 0) (c« + pP0)2 


(26) 


4 ttzMo 64 T*3f o2 

and decrement or attenuation constant: 

170. It follows, however, that under the conditions considered, 
a cumulative surging, or an oscillation with continuously increas- 
ing amplitude, can not occur, but that a synchronous motor, 
when displaced in phase from its mean position, returns thereto 
either aperiodically, if b2 > a, or with an oscillation of vanishing 
amplitude, if b2 < a. At the worst, it may oscillate with constant 
amplitude, if b = 0. 

Cumulative surging can, therefore, occur only if in the differ- 
ential equation (19): 

«»+2»2+$-0' (28) 

the coefficient, 6, is negative. 

Since c2, representing the induction motor torque of the damp- 
ing device, etc., is positive, and pPo is also positive (p being 
the exponent of power variation with speed), this presupposes 

-A2 
the existence of a third and negative term, Q rnf , in b: 

O IT J M o 

This negative term represents a power: 

P2 = -h2s; (30) 

that is, a retarding torque during slow speed, or increasing £, and 
accelerating torque during high speed, or decreasing 0. 

The source of this torque may be found external to the motor, 
or internal, in its magnetic circuit. 


SURGING OF SYNCHRONOUS MOTORS 


297 


External sources of negative, Pi, may be, for instance, the 

magnetic field of a self-exciting, direct -current generator, driven 

r the synchronous motor. With decrease of Speed, this field 

's, due to the decrease of generated voltage, and increases 

vith increase of speed. This change of field strength, however, 

i behind the exciting voltage and thus speed, that is, during 
decrease of speed the output is greater than during increase of 
speed. If this direct-current generator is the exciter of the 
synchronous motor, the effect may l>e intensified. 

The change of power input into the synchronous motor, with 
change of speed, may cause the governor to act on the prime 
mover driving the generator, which supplies power to the motor, 
and the lag of the governor behind the change of output gives a 
pulsation of the generator frequency, of e(), which acts like 
a negative power, Pj. The pulsation of impressed voltage, 
caused by the pulsation of 0, may give rise to a negative, 
/'., also. 

An internal cause of a negative term, /Js, is found in the lag 
of the synchronous motor field behind the resultant m.m.f. In 
the preceding discussion, i- is the "nominal generated e.m.f." 

I of the synchronous machine, corresponding to the field excita- 
tion. The actual magnetic flux of the machine, however, does 
not correspond to e, and thus to the field excitation, but corre- 
sponds to the resultant m.m.f. of field excitation and armature 
reaction, which latter varies in intensity and in phase during the 
oscillation of 0. Hence, while e is constant, the magnetic flux 
is not constant, but pulsates with the oscillations of the machine. 
This pulsation of the magnetic flux lags l>ehind the pulsation of 
m.m.f., and thereby gives rise to a term in 6 in equation (28). 
If PB, &, e, eu, Z are such that a retardation of the motor increases 
the magnetizing, or decreases the demagnetising force of the 
armature reaction, a negative term, P,, appears, otherwise a 
positive term. 
Pi in this case is the energy consumed by the magnetic cycle 
uf the machine at full frequency, assuming the cycle at full fre- 
quency as the same as at frequency of slip, a, 

Or inversely, e may be said to pulsate, due to the pulsation of 
armature reaction, with the same frequency as &, but with a 
phase, which may either lie lagging or leading. Lagging of the 
pulsation of e causes a negative, leading a positive, Pt, 

P~, therefore, represents the power due to the pulsation of e 


298 ELECTRICAL APPARATUS 

caused by the pulsation of the armature reaction, as discussed in 
"Theory and Calculation of Alternating-Current Phenomena." 

Any appliance increasing the area of the magnetic cycle of 
pulsation, as short-circuits around the field poles, therefore, 
increases the steadiness of a steady and increases the unsteadi- 
ness of an unsteady synchronous motor. 

In self-exciting synchronous converters, the pulsation of e is 
intensified by the pulsation of direct-current voltage caused 
thereby, and hence of excitation. 

Introducing now the term, P2 = — h*s, into the differential 
equations of paragraph 169, gives the additional cases: 

b < 0, or negative, that is : 

c2 + pPQ - h2 


8./M0 < °- (31) 


Hence, denoting: 


6l 6 . __.. , (32) 

gives: 

4. If: 6i2 > a, g = + VV - a, 

6 = ^l€ + (6l+/^ + A2€ + (6l-/)tf. (33) 

That is, without oscillation, the motor drifts out of step, in 
unstable equilibrium. 

5. If: a > 6i2, g = y/a - bS, 

8 = £€ + Mcos(00 + 6). (34) 

That is, the motor oscillates, with constantly increasing am- 
plitude, until it drops out of step. This is the typical case of 
cumulative surging by electro-mechanical resonance. 

The problem of surging of synchronous machines, and its 
elimination, thus resolves into the investigation of the coefficient: 


8x/Jlf0 


(35) 


while the frequency of surging, where such exists, is given by: 


f _ jfeeo sin (a - 0) (c2 + pP0 - /i2)2 (Wi 

Case (4), steady drifting out of step, has only rarely l>een 
observed. 

The avoidance of surging thus requires: 


SURGING OF SYNCHRONOUS MOTORS 299 

1. An elimination of the term ft2, or reduction as far as possible. 

2. A sufficiently large term, c2, or 

3. A sufficiently large term, pP0. 

(1) refers to the design of the synchronous machine and the 
system on which it operates. (2) leads to the use of electro- 
magnetic anti-surging devices, as an induction motor winding in 
the field poles, short-circuits between the poles, or around the 
poles, and (3) leads to flexible connection to a load or a mo- 
mentum, as flexible connection with a flywheel, or belt drive of 
the load. 

The conditions of steadiness are : 


and if: 


0>a, 
c2 + pP0 - h2 > 0, 

(c2 + pP0 - A2)2 ^ ee0 sin (a - 0) 

> — : 


16 t/M0 

no oscillation at all occurs, otherwise an oscillation with decreas- 
ing amplitude. 

As seen, cumulative oscillation, that is, hunting or surging, 
can occur only, if there is a source of power supply converting 
into low-frequency pulsating power, and the mechanism of con- 
version is a lag of some effect — in the magnetic field of the 
machine, or external — which causes the forces restoring the 
machine into step, to be greater than the forces which oppose the 
deviation from the position in step corresponding to the load. 
For further discussion of the phenomenon of cumulative surging, 
and of cumulative oscillations in general, see Chapter XI of 
"Theory and Calculation of Electric Circuits.,,