# 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:

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

2. UA2B2 / UA1B1 = 0.732

3. XC = XL = 3 * Un^'2 / P
where:

4. 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:

1. 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 = I
L = Ic

2. 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 I
Lmax and Icmax:
I
Lmax =  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 = I
Lmax = 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

1. 3 phase voltage ; 3 x 400V; Un = 400V

2. Frequency  : f = 50Hz

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

### Power factor correction

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

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

3. 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

1. Xc = XL = 3 * Un^2 / P  = 56,5 Ohm

2. L = XL / (2 * pi* f) =   180 mH

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

4. Ic = IL = 1.225 * Un / Xc = 8.75 A

### Balancing transformer using Tgx

1. Input voltage U11 =  400V

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

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

### Autotransformer using Tkx

1. Input voltageU11 =  400V

2. Output on the primary winding: U12= 230V, I12 = P / Uz = 8500/230 =37A

### Balanced 3 phase net

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

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

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