Phase Drop in Distribution System.

 

Phase Drop in Distribution System.

 

There may be the situation when one of the phases is not available to distribution transformer, may be due to operation DO-Fuse, Opening of jumper or purposely for load shedding etc. In this condition voltage on LV side is disturbed. 

Consider 11000/400 Volts Delta-Star distribution transformer. Out of three phase 11000 volts supply, one phase is open as shown in the figure.

There are three primary coils P1, P2 and P3 and three secondary coils S1, S2 and S3 assembled on three cores of transformer limbs A, B and C.  B is centre limb whereas A & C are side limbs as shown in figure.

                         

Case 1

When one phase say Y phase is open, primary P2 has normal line voltage V_RB across it whereas primaries P1 and P3 are in series across line voltage V_BR. The path of magnetic flux induced in the core is as shown in figure.

As seen the flux path length is equal between core B & A and between core B & C. Hence flux distribution is also equal on both the paths. Total flux in core B is divided equally in core A and C. Consequently induced voltage on secondary S2 has normal phase voltage whereas secondary S1 and S3 have half the phase voltage. Voltages in S1 & S3 are opposite to that of S2.

Rated LV Line Voltage = 400V

Therefore Phase Voltage = 400/√3 = 230V

Phase voltages

Vnb =  V_PH   = 230V

Vny = ½V_PH  = 115V

Vnr = ½V_PH  = 115V

 Line Voltages

Vby = V_PH  + ½V_PH  = 230 + 115 =345V                        Addition of In line voltages.

Vbr = V_PH  + ½V_PH  = 230 + 115 =345V                         Addition of In line voltages.

Vyr = 00

Case 2.

When one phase say B phase is open, primary P1 has normal line voltage V_RB across it whereas primaries P2 and P3 are in series across line voltage V_BR. The path of magnetic flux induced in the core is as shown in figure.

As seen the flux path length is unequal between core A & B and between core A & C. Hence flux distribution is also unequal on both the paths. Total flux in core A is divided unequally in core B and C. Consequently induced voltage on secondary S1 has normal phase voltage whereas secondary S2 has more than half the phase voltage and S3 has less than half the phase voltage. Voltages in S2 & S3 are opposite to that of S2.

Rated Line Voltage = 400V

Therefore Phase Voltage = 400/√3 = 230V

Let dv is 30V, it depends on core structure.

Phase Voltages

V_PH = 230 V

Vnr =  V_PH  = 230 V

Vny = ½V_PH + dV = 115 - 30 = 85V                                     Addition of In line voltages.

Vnb = ½V_PH  - dV = 115 + 30 = 145V                                  Addition of In line voltages.

Line Voltages

Vry = V_PH  + ½V_PH  - dV = 230 + 115 - 30 = 315V               Addition of In line voltages.

Vrb = V_PH  + ½V_PH + dV = 230 + 115 + 30 = 375V             Addition of In line voltages.

Vyb = 2dV = 2 × 30 = 60V

Secondary voltages are not normal phase voltage. One phase has normal voltage whereas other two phases have half or little more or less than half voltages

Secondary voltages are in line. There is no phase shift between voltages. 

Therefore this supply is not suitable to produce balanced rotating magnetic field to run motor. Hence suitable to suppress motive power load in distribution system and continue essential single phase loads for light, fan, refrigerators etc. This phenomenon is used for load shedding in distribution system for load control. All single phase consumers are kept on phase getting normal voltages.

Some mischievous users use voltage booster and capacitor to get normal voltage and phase shift on other phases and run three phase motor. But it is risky for User as well Supply Company in view of probability of failure of motor and/or distribution transformer.