Full resolution (JPEG) - On this page / på denna sida - 1958, H. 9 - The Balancing of Unbalanced Three-Phase Loads, by Sune Rusck
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stant frequency. A primary three-phase
transformator T, fig. 1, is connected to this network. The active
and reactive losses of this transformer will be
neglected. This latter assumption will not in any way
impose any limitations on the following general
treatment, since the transformer losses can always
be included in the load. The loads are connected to
the n secondary terminals of the transformer. The
power taken out in the secondary phase v is
consequently Sv = Pv + j Qv. The actual problem can
now be formulated in the following manner. What
relationship exists between, on the one hand, the
primary load taken out and, on the other hand, the
secondary loads and the phase angles of the
secondary voltages when the load on the primary side is
balanced?
The condition for a balanced three-phase load
From the theory for the symmetrical components it
is well known that an arbitrary unbalanced
three-phase current can be divided up into two
symmetrical three-phase currents, of which the one, the
posi-tive-sequence current i+, has the phase sequence RST,
and the other, the negative-sequence current /_,
has the phase sequence RTS. (We assume that there
is no zero-sequence current.) The instantaneous
values for the primary load currents ir, is and i t of
the transformer can consequently be written as
iR = \/2 i+ eos (cot + ßi) + VJi_ eos (cot + ß2)
is = V2 i+ eos [cot + ßi- + Y2i_cos[ot + ß2 +
IT
= V2 i+ eos (cot + ßi + ^-) + l/2~i_cos(wf +
If we multiply equation (1) with the expressions
for the instantaneous values of the symmetrical phase
voltages
ur = V^u eos (cot + ßi + cp)
us = ] 2~u eos [at + ßi + cp —
iit = V2 u eos [
2jt\
cot + ßi + cp + —J
(2)
and add the result, the instantaneous power taken
out from the three-phase network p = uRiR + usis +
+ uyi’r is obtained as
p = 3 u i+ eos cp + 3 u i_ eos (2 o>t + ßi + ß2 + cp) (3)
The instantaneous power thus comprises a constant
part, determined by the positive-sequence current
and its power factor, and of a part varying
sinus-oidally with the double network frequency
determined by the magnitude of the negative
phase-se-quence current. If = 0, the three-phase currents
are fully symmetrical according to equation (1). From
equation (3) it follows then that, if the
instantaneous three-phase power is constant, the three-phase
load is balanced. This relationship will be applied
when formulating the general balance condition.
We shall now seek the relationship between the
loads taken out on the secondary side and the power
consumed on the primary side. It is assumed that
the instantaneous voltage across the winding v is
]/2 uv eos (o)t + ocv )and the instantaneous current in
the same winding is j/2 iv eos (cot + ocv — cpv), where
ocv is the phase angle of the voltage and <pv the phase
angle of the load.
If we multiply the two expressions with one another
and summate the results across all the n secondary
windings, the sum of the instantaneous power taken
out on the secondary side is obtained. Since it is
assumed that the transformer is free from losses,
this sum will be equal to the instantaneous power
applied to the primary side. After certain
simplifications, we obtain the following equation
(1)
Fig. 2. Graphical solutions of the balance condition
equation (7)
a) The general case, n = 4.
b) The optimum case, n = 2.
c) The optimum case, n = 3.
1 122 ELTEKN I K 1958
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