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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primary side will be
P + ]Q = Pi + KQi + O2 + Qd = Pi + 3/Qi (10)
The power factor on the primary side is
consequently obtained as
eos (p =
Pi
COS (pl
V P2i + 9 Q2i |/9 —8 eos2 <p\
(11)
In fig. 3, curve a, the primary power factor is
shown as a function of the power factor of the
single-phase load. From the figure it is clearly
apparent that this balancing method is very unfavourable
with regard to the power factor of the primary side.
A power factor of eos <px = 0.8—0.9 for the
single-phase load will consequently give on the primary
side only eos <p = 0.49—0.67. This value can
naturally be improved as a result of phase
compensation in such a manner that Qt is compensated with a
reactive load — Qx connected to the same phase as
the single-phase load Px + j In this case, we
obtain on the primary side
P + jQ = Pi + j 3 (Q, - Q.r) (12)
The phase compensation can also be carried out on
the primary side, but this method demands a larger
amount of installed reactive volt-amperes.
As an indication of the costs involved when
balancing, we can use the total installed reactive
volt-amperes divided by the active power. If we call
this value for K, we obtain
K =
+ [Q^l + 1 Q,
Pi
(13)
In fig. 4, K has been plotted as a function of the
power factor eos <px of the load. The curve Kx refers
to the case where Qx — — Qi, i.e. eos <p = 1, whereas
curve K2 applies to Qx = 0. The two curves denote
the limits within which K can vary in practice. In
Fig. A. The installed reactive volt-amperes as a function
of the secondary power factor.
order to balance a single-phase load with the power
factor 0.8, reactive volt-amperes of 1.5—1.8 times
the active power is consequently required. The
corresponding value would have been 1.25 for the
optimum case.
Two-pliase balancing
We shall now investigate the case where the
balancing is carried out in the optimum manner with only
one reactive load. The following solutions of
equation (7) are obtained
Q2 = ± jfptx +
_ 71 (p
ct2 = +–—
4 2
(14)
Out of consideration for the power factor on the
primary side, it is desirable to select for Q2 the
opposite sign to that for Qx. Thus, if Q± is inductive
Q2 = — \ P\ + Q\ and oc2 = ji/4 — *p/2 will apply.
Fig. 5 shows how this case can be achieved with the
aid of an auto-connected transformer. The
three-phase power will be
p + jq = Pi - j (yp21 + q2x
i.e
____ we have a reactive input
The power factor is
Q1) (15)
on the primary side.
eos (p = —=r
eos (p\
V2 Fi
sin <p 1
(16)
This function has been plotted in fig. 3 as curve b.
If we disregard the values immediately below
eos <px — 1, it can be seen that high values are
obtained for the power factor.
The reactive volt-amperes required for balancing
divided by the active power will be in this case
K
1
COS (p\
(17)
This curve is reproduced in fig. 4 and is marked
K3. As can be seen, this optimum manner of
balancing a single-phase load will require considerably
less reactive volt-amperes than if balancing takes
place on three phases. When making an economic
comparison, it is necessary to take into
consideration, however, the costs involved when producing
the special phase angle for the balancing voltage
i.e. the transformer shown in fig. 5. This
transformer may, however, be included in the step down
transformer.
Unbalance with a varying load
The balancing can naturally only be precise for a
single value of the load, which means that a
negative-sequence current will be obtained on the
primary side for a varying load. If no measures for
balancing are taken, the negative phase-sequency
current will be according to equations (3) and (4)
VP2I + Q2
3 u
IM
3 u
(18)
If the single-phase load is balanced for the value
Sx = Pt -f j Q± and the load is temporarily increased
to 5, + A = P, + A Pl + j (Öl + A öl) a negative
phase-sequence current is obtained in an analogous
manner.
1 124 ELTEKN I K 1958
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