Three phase unbalanced load flow 2

[EddyWirbelstrom]

A balanced 240V a.c. phase-to-phase three phase ( A-B-C ) star connected source of equal negligible positive and negative sequence impedance and no connection to earth (ground) supplies single phase 240V a.c. incandescent lamps.
Each lamp is connected between 240V phase-to-phase at 5 m intervals along a three core ( A-B-C ) cable with the objective of achieving three balanced line cable currents ( A-B-C ) . Eg. lamp connections A-B, B-C, C-A .....
Can anyone suggest how the load flow currents and can be calculated

 

[RRaghunath]

The incandescent lamps in the said case are connected in Delta fashion and are said to be presenting balanced load among all three phases.
Hence, the current in A-B-C lines will be that in any of the three circuits (same current as the loads are balanced) multiplied by sqrt3.

 

[che12345]

Dear Mr. EddyWirbelstrom (Electrical)(OP)19 Sep 22 11:18
" ... A balanced 240V a.c. phase-to-phase three phase ( A-B-C ) star connected source.... no connection to earth .... supplies single phase 240V a.c. incandescent lamps. Each lamp is connected between 240V phase-to-phase....( A-B-C ) cable ...achieving three balanced line cable currents ( A-B-C ) . Eg. lamp connections A-B, B-C, C-A .....Can anyone suggest how the load flow currents and can be calculated "
1. Your source ABC is in Y formation. Neutral is non-earthed. VAB =VBC =VCA =240 V .
1.1 Attention: Take note that VAB , VBC, VCA are 120deg apart.
2. Your 240V lamps are connected in D formation, across AB, BC, CA.
3. You NEED to know the wattage of the lamps in order to calculate the current through each lamp.
3.1. The Current through each lamp will be the same=balance = i =W/V, where V=240V , W=wattage W.
4.1. The current from the source will the same=balance IA=IB=IC ... A.
5. The source current IA=IB=IC = 1.732 x current through each lamp, see above 3.1.
Che Kuan Yau (Singapore)

 

[jghrist]

If the phases are not balanced, you can calculate the current from B-A, A-C, and C-B vectorially, knowing the Ø-Ø voltage and the resistance of the lamps (V²/P where V is rated voltage and P is rated watts). IBA = VBA/R, IAC = VAC/R, ICB = VCB/R. Then calculate the line current vectorially IA = IAC - IBA, IB = IBA - ICB, IC = ICB - IAC.

 

[7anoter4]

If we neglect the supply cable impedance and consider sum of all lighting lamps per each phase, we get a triangle of unbalance resistances. If we transform this triangle into a star, we get the currents per phase. The voltages are equals 240/sqrt(3).

 

[jghrist]

Zanoter4,

OK except that there is no neutral, so the phase voltages are not defined. If you assume balanced 3Ø voltages, then if the loads are unbalanced, the phase currents will not sum to zero. They have to aum to zero because there is no neutral. See the attached Excel workbook.

 

[7anoter4]

Thank you very much, jghrist. Because the neutral is not in the centre VAN,VBN and VCN are not equal to the phase voltages and then the sum of the currents is not equal to zero. It is necessary to calculate the potential of the neutral point first.

 

[7anoter4]

I think I get it

 

[FacEngrPE]

The phase voltages of an ungrounded Y are defined with respect to each other, but not ground. For this problem all of the loads are between phases so the relationship of ground to any phase does not matter.

If it is convenient to choose to pick a point in the center of the Y for the purpose of calculation, the fact that it is not defined with respect to ground does not matter either.

Caution - sometimes the ground reference matters, it depends on the question.

 

[7anoter4]

Thank you, FacEngrPE . In the open post it is written "star connected source”. New VN (load neutral point potential) it is the voltage referred to the source star connection point -grounded or not.