CHAPTER XVI 

LOAD BALANCE OF POLYPHASE SYSTEMS 

163. The total flow of power of a balanced symmetrical poly- 
phase system is constant. That is, the sum of the instantaneous 
values of power of all the phases is constant throughout the cycle. 
In the single-phase system, however, or in a polyphase system with 
unbalanced load, that is, a system in which the different phases 
are unequally loaded, the total flow of power is pulsating, with 
double frequency. To balance an unbalanced polyphase system 
thus requires a storage of energy, hence can not be done by 
any method of connection or transformation. Thus mechanical 
momentum acts as energy-storing device in the use as phase bal- 
ancer, of the induction or the synchronous machine. Electrically, 
energy is stored by inductance and by capacity. The question 
then arises, whether by the use of a reactor, or a condenser, con- 
nected to a suitable phase of the system, an unequally loaded 
polyphase system can be balanced, so as to give constant power 
during the cycle. 

In interlinked polyphase circuits, such as the three-phase sys- 
tem, with unbalanced load carried over lines of appreciable im- 
pedance, the voltages of the three phases become unequal. This 
makes voltage regulation more complicated than in a balanced 
system. A great unbalancing of the load, such as produced by 
operating a heavy single-phase load, as a single-phase railway or 
electric furnace, greatly reduces the power capacity of lines, trans- 
formers and generators. Unbalanced load on the generators 
causes a pulsating armature reaction: at single-phase load, the 
armature reaction pulsates between more than twice the average 
value, and a small reversed value, between f (cos a + 1) and 
F(cos a — 1), where cos a is the power-factor of the single-phase 
load. Especially in alternators of very high armature reaction, 
as modern steam-turbine alternators, a pulsation of the armatiu^ 
reaction is very objectionable. It causes a pulsation of the field 
flux, leading to excessive eddy-current losses and consequent re- 
duction of the output. The use of a squirrel-cage winding in the 

314 


LOAD BALANCE OF POLYPHASE SYSTEMS 315 

field pole faces of the single-phase alternator reduces the pulsation 
of the field flux, but also increases the momentary short-circuit 
stresses. 

Thus, it is of interest to study the question of balancing unbal- 
anced polyphase circuits by stationary energy-storing devices, as 
reactor or condenser. 

164. Let a voltage, 

e = E cos <l> (1) 

be impressed upon a non-inductive load, giving the current 

i = I cos (2) 


The power then is 


where 


p = ei = EI cos^ 
= ^ (1 + cos 2 <t>) 
= Q + Q cos 2 « (3) 

= f (4) 

that is, 

in a non-inductive single-phase circuit, the power consists of a 

constant component, 

Q--2' 

and an alternating component, 

EI 

= "2- cos 2 0, 

of twice the frequency of the supply voltage, and a maximum 
value equal to that of the constant component. The instantane- 
ous power thus pulsates between zero and 2 Q, by equation (3). 
If the circuit is inductive, of lag angle a, the current is 

i = I cos (0 — a) (5) 

and the instantaneous power thus, 

p — EI cos 4> cos (0 — a) 

= -rt" cos a + cos (2 — a) 

= P -h Q cos (2 - a), 
thus consists of a constant component, 

P = -2- cos a = Q cos a (7) 


316 ELECTRIC CIRCUITS 

and an alternating component, 

Q cos (2 — a) ; 

it thus pulsates between a small negative and a large positive 
value, P - Q and P + Q. 

If the circuit is completely inductive, that is, the current lags 

IT 

90° or 2 behind the voltage, the current is 

i = / cos (« - I) (8) 

and the instantaneous power thus, 

p = EI cos <l> cos(<^ — ^j 

= -^ sm 2 

= Qcos(20-^)i (9) 

Thus, the power comprises only an alternating component, but 
no continuous component; in other words, no power is consumed, 
but the power surges or alternates between +Q and — Q, that is, 
power is stored and then again returned to the circuit. 

If the circuit is closed by a capacity, C, the current leads the 

TT 

impressed voltage by ^, thus is 

i = / cos (« + I) (10) 

and the instantaneous power thus, 

p = EI cos <f) cos (0 + 9) 

= Qcos(2<^ + |) (11) 

thus, comprises only an alternating component, surging be- 
tween — Q and +Q, with double frequency. 

The power consumed by a condenser, equation (11), is opposite 
in sign and thus in direction, from that consumed by a reactor (9), 

Qcos(2<t> + l) = -Q cos(2 « - ^) • 

166. If a number of voltages, 

ei = Ei cos (<^ — 7i) (12) 

* "Engineering Mathematics," Chapter III, paragraphs 66 to 75. 


LOAD BALANCE OF POLYPHASE SYSTEMS 317 

of a polyphase system, produce currents, 

ii = li cos (<t> — yi - ai) (13) 

the instantaneous power of each voltage e* is 

Pi = edi 

= Q<{cos ai + cos (2 - 2 7i - ai) ] (14) 

and the total instantaneous power of the system thus is 

V = 2)pi 

= SQ< cos ai + SQi cos (2 <^ — 2 7* — «») 
= P + Q cos (2 <^ - a) (15) 

where 

P = SQiCOStti (16) 

is the total effective power of the system, and 

Q = SQi cos (2 - 2 7i - «») (17) 

is the total resultant alternating component of power, or the 
resultant power pulsation of the system. 

Thus, the power of the polyphase system pulsates, with double 
frequency, between P — Q and P + Q. 

In this case, P may be greater than Q, and frequently is, and the 
power thus pulsates between two positive values, while in the 
single-phase circuit (6) it pulsated between positive and negative 
value. 

It thus can be seen, that in any system, polyphase or single- 
phase, with any kind of load, the total instantaneous power of the 
system can be expressed, 

p = P H- Q cos (2 - a) (18) 

where P is the constant component of power, and Q the amplitude 
of the double-frequency alternating component of power, and Q 
may be larger or smaller than P. 

It must be noted, that Q is not the total reactive power of the 
system — which would have to be considered, for instance, in 
power-factor compensation etc. — but Q is the vector resultant 
of the reactive powers of the individual circuits, while the total 
reactive power of the system is the algebraic sum of the individual 
reactive powers (see "Theory and Calculation of Alternating- 
current Phenomena," Chapter XVI). 

Thus, for instance, in a system of balanced load, even if the 
load is reactive, Q = 0. Thus, Q is the unbalanced reactive 


318 ELECTRIC CIRCUITS 

power of the system, and does not include the reactive power, 
which is balanced between the phases and thereby gives zero 
as vector resultant. 

166. The expression of the power of a polyphase system of gen- 
eral unbalanced load is by (15) 

p = P + Q cos (2 <^ - a) (19) 

this also is the expression of power of the single-phase load of 
lag angle a, of the impressed voltage and current, 

e = E cos <t> 
1=1 cos (<p — a) 

where, from (20), 

P = Q sin a 
EI 


(20) 


Q = 


(21) 


2 

while in the general case (19) P and Q may have any values. 
Suppose now we select from the polyphase system a voltage, 

e' = E' cos (« - p) (22) 

and load it with an inductive load of zero power-factor, 

i'-rcos(«-i8-^) (23) 

E' . 
that is, we connect a reactor of a; = -y> into the phase e'. 

The power of (22) (23) then is 

p' = Q'cos(2<^-2<8-|) (24) 

where 

Q' = ^ (26) 

and the total power of the system, comprising (19) and (25), thus 
is 

Po = p + p' 

= P + Q cos (2 <^ - a) H- Q' cos ^2 <^ - 2 18 - ^) 

and this would become constant, and the double-frequency term 
eliminated, that is, the system would be balanced, if Q' and P are 
chosen so that 

Q cos (2 - a) + Q' cos (2 « - 2/3 - I) = (26) 


LOAD BALANCE OF POLYPHASE SYSTEMS 319 

hence, 

& = Q (27) 

2 - 2 18 - I = 2 - a - IT 


or, 


thus. 


/3 = f + I (28) 

2 ^ 
?' = ^ (29) 

e' = E' cos [0 - (I + I) ] (31) 

is the voltage, which, impressed upon a reactor of reactance, 

x = §Q (30) 

balances the power, 

P = P + cos .(2 - a) (24) 

of an unbalanced polyphase system. That is, 

e' = E' cos [* - (f + J) ] (31) 

impressed upon the reactance, x, gives the current, 

'■-^-[*-(i+T)] »^' 

and thus the power, 

p'-«eo.[*-(| + |)]cos[*-(J + %')] 

= - e COS (2 <^ - a) (33) 

and this reactive power, p', added to the unbalanced polyphase 
power, p, gives the balanced power, 

p = p + p' 
= P. 

167. Comparing (31) with (20) or (24), it follows: 
The unbalanced load of a single-phase voltage, 

ft 

e = E cos 0, 


320 ELECTRIC CIRCUITS 

of lag angle, a, or in general, the unbalanced load of a polyphase 
system with the resultant instantaneous power of lag angle, a, 

p = P + Q cos (2 « - a) 

can be balanced by a wattless reactive load, p', having the same 
volt-amperes, Q', as the alternating component, Q, of the imbal- 
anced load, and having a phase of voltage lagging by 

a .T 
2 "^4 

or by 45° plus half the lag angle, a, of the unbalanced load or un- 
balanced single-phase current. 

Just as the unbalanced polyphase load, p, (24) may be single- 
phase load on one phase, or the vector resultant of the loads on 
different phases, so the wattless reactive compensating volt- 
amperes (33) may be due to a single reactor connected into the 
compensating voltage, e', (31), or may be the vector resultant of 
several voltages, e'l, loaded by reactances, Xi, so that their 
vector resultant is p' (33). 

If a capacity is used for energy storage in balancing unbalanced 
load (24), the compensating voltage (22), 

e' = E' cos (<^ - j8), 

impressed upon the capacity gives the reactive leading current, 

2' =rcos(«-i8 + |) (34) 

hence the compensating reactive power, 

p' = E'rco8(2<t>-2fi + ^ (35) 

and therefrom, by the same reasoning as before, 

^ = 1+^ (36) 


e' = 4cos<^-(|+^)] 


(37) 


That is, when using a capacity for balancing the load, the com- 
pensating voltage, e', has the phase, 

a, 3 TT 


LOAD BALANCE OF POLYPHASE SYSTEMS 321 


or, what is the same as regards to the power expression, 

a TT 

2-4' 

thus lags by half the phase angle, a, minus 45° (or plus 135°). 

168. As instance may be considered a quarter-phase system 
with one phase loaded. 

Let 

d = E cos <t> 


€2 = E cos 


(*-5) 


(38) 


be the two phase-voltages of the quarter-phase system. 

Let the first phase, ei, be loaded by a current lagging by phase 
angle, a, 

2*1 = / cos (<^ — a) (39) 

while the second phase, €2, is not loaded. 
The power then is 


V = eiii 

EI 

= -^{cos a + cos(2<^ — a)} 


(40) 


and is compensated or balanced by a reactance connected to a 
compensating phase, 

e' = E' cos {<l>- p) (41) 

and consuming the reactive current, 


i' = r cos[<t> - p + 1) 


(42) 


TT 


T 


where the — ^ represents inductive reactance, the + « capacity 

reactance. 

The compensating reactive power then is 


p' = e'i' 


'T' 


E'l 


and this becomes equal to 


cos(2<^-2/3T0 


(43) 


for 


2" c^s (2 * - a), 


ET = EI 

^-2+4 


21 


(44) 


322 


ELECTRIC CIRCUITS 


and the compensating circuit thus is 

e' = i;'co8(0-|T?) 

•/ T/ / . a _^ 3 w\ 


it is, then, 


hence. 


for 


= £7' cos (2 <^ - a HF ^) 
= -EI cos (2 - a) 

Po = p + p' 

EI 
= -^ cos a, 

a = 0, or non-inductive load, it is 
e' = E' cos (« T ^) 

= ^jcos«±cos(«-^) 


hsnce, if we choose, 
hence, 


E' = E \/2, 


it is 


/' = 


^' 


e = Ci ± ej 


(45) 


(46) 


that is, connecting the two phases in series, gives the compensat- 
ing voltage for non-inductive load. Or: 

"Non-inductive single-phase load, on one phase of a quarter- 
phase system, can be balanced by connecting a reactance across 
both phases in series, of such value as to consume a current equal 
to the single-phase load current divided by \/2, that is, having the 
same volt-ampere as the single-phase load." 

169. In the general case of inductive load of power-factor, a, 
the compensating voltage (45) can be written, 

e' =^ W cos(^ ± tJc^s 4> + sin(^ ± 7) sin <^ 


= E' 


cos^l ± ^)cos <^ ± cos^^ + 5)^0^(0 - I) J, 


LOAD BALANCE OF POLYPHASE SYSTEMS 323 


or, choosing, 


E' = £?, 


thus, 


/' = /, 


it is, by (38), 


1 


e' 


= Oi^i ± O262 


where 


Ol 


= cos (1 ± ?) 


a2 


= COS (1 + 1) 


(47) 


and the upper sign applies to the reactor, the lower to the con- 
denser as compensating circuit. 
The current then is 

t' = /cos(*-|T^)- (48) 

The compensating voltage e' thus can be produced by connect- 
ing a transformer of ratio Oi into the first phase, ei, a transformer 
of ratio, aj, into the second phase, 62, and connecting their second- 
aries in series across a reactor or condenser of suitable reactance. 

The current, i\ in the compensating circuit consumes a current, 
oif, in the first phase, Ci, and a current, 021', in the second phase, 
62. As the latter phase has no load, the total current in the 
second phase is 

t2 = 02t' = / cos(2 T ^J cos \^* - 2 ^ T/ 

the total current in the first phase is 
ii® = ii + dii' 

= /jcos {<!> - a) + cos (I ± ^) cos (* - I HF -/) 

= / cos(<^ — a) + 0.5 cos (* + I) + 0.5 cos {<!> — a ^ t) 

= 0.5 /] cos (<^ — a) + cos ((j) hF ^) 

-'H-^i)-(*-l*Dl' 

hence has the same value as ^2, but differs from it by k ^^ ^^^ ^^ 
phase, thus has to its voltage, Ci, the same phase relation as ia 


324 ELECTRIC CIRCUITS 

has to its voltage, d. That is, the system is balanced in load, in 
phase and in armature reaction. 
In the unbalanced single-phase load, the power-factor is 

tti = cos a 

in the balanced load, the power-factor is 

ax = cos (I + ^) 
thus, is materially reduced for a reactor as compensator, +-', 

it is in general increased for a condenser as compensator, —z- 

170. Instead of varying the phase angle of the compensating 
voltage, e', with varying phase angle, a, of the single-phase load, 
compensation can be produced by compensating voltages of 
constant-phase angle, utilizing two such voltages and varying the 
proportions of their reactive currents, with changes of a. 

Thus, if 

ii = / cos (<^ — a), 
is the load on phase, 

ei — E cos <py 
and the second phase 

62 = E cos (* - ^) > 

is not loaded, thus giving the unbalanced power, 

EI 

p = -^{ cos a + cos (2 <^ — a) } (49) 

as compensating voltage may be used, 

the voltage of both phases connected in series, 

e = ei + e^ 
= ^\/2cos(*-^) (50) 

and the voltage of the second phase, 

e2 = ^cos(«-|)- (51) 

Let, then, 

i' = /' cos(<^ j-)> 

be the reactive current of the compensating phase, e, and 

i'2 = 1^2 cos {<t> — tt), 


LOAD BALANCE OF POLYPHASE SYSTEMS 325 


= — /'2 COS <l> 

the reactive current of the compensating phase, d. 
The powers of the two compensating circuits then are 


p = ei 


_ Ery/2 


2 

Ery/2 


cos (2 <t> — tt) 


cos 2 <t> 


(62) 


and 


p'2 = e2i'2 


EP. 


cos 


(^*-3 


(53) 


and the condition of compensation thus is 

— cos (2 <^ - a) = — cos2 <^ + -^ cos^^2 <t> - 2) ^^^ 

or, resolved, 

(/ cos a — P\/2) cos 2 <^ + (/ sin a — P2) sin 2 <^ = 0, 

and as this must be an identity, the individual coefficients must 

vanish, that is, 

/ cos a 


V2 


P2 = / sin a = / cos (a -- ^j 


(55) 


thus, the compensating voltages and currents, which balance the 
single-phase load, 

ei == E cos <f> 

ii = / cos (<^ — a) 


(56) 


are 


e = ei + 62 
= E\/2 cos 

/ cos a 
i = ^^^cosU 


and 


(-0 


(57) 


ei= E cos ( <^ — |j 

i'i — — /cos (a — ^jcos^ 
= / sin a cos ^ 


(58) 


326 


ELECTRIC CIRCUITS 


As seen, this means loading the second phase with a reactor 

giving the same volt-amperes, 

EI . 
-^ sm a, 


e2i'2 = 


as the unbalanced single-phase load (56), and thereby balancing 
the reactive component of load, and then balancing the energy 
component of the load by the compensating voltage ei + ej, as 
given by (46). 

If the single-phase load is connected across both phases of 
the quarter-phase machine in series, 

€ = 61 + 62 

= EV2 cos(«T^) 


t = ;^cos(«4-^-a) 


(59) 


in the same manner the conditions of compensation can be de- 
rived, and give the compensating circuit, 


e' = E' cos U - I) 
i' = /'cos(«-|-^) 


(60) 


where 


ET 


= EI. 


For non 


-inductive load, 


a 


= 0, 


this gives 


e' 


= Ci, 


that is, one of the two phases is compensating phase for the re- 
sultant. 

171. As further instance may be considered the balancing of 
single-phase load on one phase of a three-phase system. 
Let 


61 = E cos 9 

62 = E cos(<t> ^) 


63 = E cos (0 - -g^j 
be the three voltages between the three lines and the neutraL 


(61) 


LOAD BALANCE OF POLYPHASE SYSTEMS 327 


The voltage from line 1 to line 2, then, is 

Ci* = 61 — 62 

= EVs cos (<A + 


(62) 


and if a = lag of current behind the voltage, the current produced 
by voltage, e, is 

i = / cos (<^ + ^ — aj , 

thus the power, 

ElV^l 


V = 


cos a + cos ( 2 <^ + - — a] 


(63) 


and this is balanced by the compensating voltage and current, as 
discussed before. 


it is 


thus, 


p' = e't' 

Po = p + p' 
EIV3 


cos a. 


thus balanced. 

The balancing voltage (64), 

e' = £rV3co8(«-|-^), 
lags behind the load voltage, e (62), by 

2+4' 

or by half the lag angle of the load, plus 45°. 
U 

"°6' 


(64) 


(65) 


328 ELECTRIC CIRCUITS 

or 30^ lag, it is 

e' = JS?V3cos(<^ -g) (66) 

thus the compensating voltage, e', is displaced in phase from the 
load voltage, eu (62), by 60°, if the lag angle of the load is 30**, and 
in this case, the second phase of the three-phase system thus can 
be used as compensating voltage, 

en = ei — ez 

= £\/3cos(<^-g 

= 6'. 

In the general case, for any lag angle, a, the compensating vol- 
tage (64) can be produced by the combination of the two-phase 
voltages, ei and ez, as 

e' = aiei — 02^3 

similar as was discussed in the quarter-phase system. 

The second phase, en, as compensating voltage, loaded by a re- 

actor, balances the load of phase angle, a = ^> or 30°. For other 

angles of lag, either another phase angle of the balancing voltage 
is necessary, or, if using the same balancing voltage, the balance 
is incomplete. 

Let thus: 
the load 

612 = EV3cos(^it> + '^, 

i = / cos ( 4- ^ - «j , 
be balanced by reactive load on the second phase, 

613 = EV^ cos (* - I) » 
i' = / cos ( ^J , 


it is: 

power of the load. 


V = eui 

EI\/3\ 


balancing power. 


I cos a -t- cos (2 4- ^ — a) [ ; 


EIVS 


cos 


(^*+i) 


LOAD BALANCE OF POLYPHASE SYSTEMS 329 


thus, total power, 
Po = p + p' 

g/Vs 

2 
ElVl 


cos a + cos (20 + 


I"")- 


COS (2<t> + 


I) 


and 


cos 


3 = 


sin(|-«) 


COS a 


is the ratio of the remaining alternating component of power, 
to the constant power, and may be called the coefficient of 
unbalancing.