Balancing a 3 Phase Net
with a Single Phase Load
Using a Single Phase Transformer and a LC Tank
General Information
Using this circuit you can distribute (balance) a
single phase load in a three phase net as follows:
P1 = P2 = P3 = S cos(φ) / 3
where:

S => Nominal power of the
single phase load. It must not vary during the operation

φ => Phase angle of the single
phase load. It must be in the range between 23° and +24°. If iit
is not in this range then you have to compensate it to optimal value φ =
0 ( Z = R).

P1, P2, P3 => Balanced power
per phase. Note that the power factosr of these balanced powers are
always equal 1 and they do not depend on the power factor of the
single phase load.
Main propertiese
For an optimal balancing of the single phase load with
this circuit you have to have the following conditions:

The thre
phase line voltages has to be balances: U12 = U23 = U31 = Un

UA2B2
/ UA1B1 = 0.732

XC
= XL = 3 * Un^'2 / P
where:
 Un => Line voltage
P => Active power of the compensated single phase load
XL = L * ω and Xc = 1 /(C * ω)
There are two ways to use this circuit:

You know the
nominal voltage U, the nominal power S and the power factor of the
single phase load. After you have compensated the single phase load you
know the active power P of the single phase load .
The transformer secondary winding has only the center tap.
Using the vector diagram for φ = 0 you can calculate the value of the
voltages Uc and UL: Uc = UL =
1.2225 * Un and the rrms value of the currents IL
and Ic: IL = Ic = UL
/ XL = Uc / Xc
For the transformrr design use the Small Transformers Program:
Input voltage : U11= Un
Output 1 in autotransformer connection; U12 = U and I12 = Iz = P / U
Output 2 : U2 = 0.732 * Un and I2 = IL
= Ic

You are going
to design a universal balancing equipment for:
 3 phase voltages 3 x Un
3 single phase output voltages Uz1, Uz2 and Uz3
 3 output powers per each single phase output S50%,
S75% and S100%
= S
 any phase angle of the single phase load between
23° and +23°
Using the vector diagram for φ = 23° and φ =+23° you can
calculate the maximum value of the voltages Ucmax and ULmax:
Ucmax = ULmax = 1.506 * Un
and the rms value of the currents ILmax
and Icmax:
ILmax = Icmax = ULmax
/ XL = Ucmax / Xc
In order to enable the balabNcing at the all single phase loads S50%, S75%
and S100% you have to use 3 parallel connected capacitors C25%%,+
C25% and C50%
 At the single phase load S100%% C = C25% + C25%
+ C50%
 At the single phase load S75% C = C25% + C50%. Non used C25% is
connected parallel to the inductor L
 At the single phase load S50% C = C50%. Two non used C25% are
connected parallel to the inductor L
For
AT the transformrr design use the Small Transformers Program:

Input voltages : U11= Uz1, U12= Uz2 and U13= Uz3

Output 1 in autotransformer connection; U14 = Un and I14 =S100'% / Un

Output 2 : U2 = 0.732 * Un and I2 = ILmax
= Icmax
The transformer secondary winding has to have more taps. The
center tap and 2 x 4 taps
should be good enough for a fine tuning of the balancing..
Design Example
Note tjat in this design example the AT
transformer is exchanged with the balancing transformer and the
autotransformer
Input

3 phase voltage ; 3 x 400V; Un =
400V

Frequency : f = 50Hz

Single pjase load: S = 10kVA,
power factor = o.85 inductive , U = 230V
Power factor correction

Active power of the single phase
load: P = 0.85 * 10000 = 8500W

Inductive power of the single
phase load : Q = ( S^2  P^2 )^0.5 = 5267VAr

Compensation capacitor connected
on the voltage U = 230:
Ccomp = Q / ( 2 * pi * U^2) = 317 uF for nominal voltage 250V
Inductor L and Capacitor C

Xc = XL =
3 * Un^2 / P = 56,5 Ohm

L = XL
/ (2 * pi* f) = 180 mH

C = 1 / Xc / (2 * pi * f) = 56
uF for nominal voltage 600V

Ic = IL = 1.225 * Un / Xc = 8.75
A
Balancing transformer using Tgx

Input voltage U11 = 400V

Secondary 1;
U21=146V, I21 = Ic = 8.75 A, angle = 45°

Secondary 2;
U22=146V, I22 = IL = 8.75 A, angle = 315°

U2 = 293 V , I2 = 8.75
A
Autotransformer using Tkx

Input voltageU11 = 400V

Output on the primary winding:
U12= 230V, I12 = P / Uz = 8500/230 =37A
Balanced 3 phase net

Phase power P1 = P2 = P3 =
P / 3 = 8500 / 3 = 2833 W

Phase current I1 = I2 = I3
= P1 / U = 2833 / 120 = 12.31 A

Power factor cos(θ1)
= cos(θ2) = cos(θ3)
= 1
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