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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
4. where:

5. Un => Line voltage
6. 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.
2. 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

3. You are going to design a universal balancing equipment for:
4. - 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 that 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 phase 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:
4. 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°
4. U2 = 293 V , I2 = 8.75 A

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