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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.
Balancing a 3 Phase Net with a Single Phase Load Using a Single Phase Transformer and a LC Tank Balancing a 3 Phase Net with a Single Phase Load Using a Single Phase Transformer and a LC Tank

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:

  4. Un => Line voltage

    5. 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

  2. You are going to design a universal balancing equipment for:

    3. - 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
    3. Frequency: f = 50Hz
    4. 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