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Votlage Drop for Unbalanced three phase

Hi all,


Hoping someone can help so i can forget about this and get to a beer garden..


For a balanced three phase voltage drop, the three or four core mv/a/m figure can be used from the releavnt table to determine the voltage drop.


However, for an unbalacned system ie a three phase submain feeding a DB, which supplies single phase loads gets more confusing. After reading previous posts, it seems the best way to deal with this is to calculate the single phase drop from each phase to the final circuit. However, this is where i am finding conflicting infromation, calculations for electricians and designers states that:

"By inspection of the extract from Table 4D4B below, it can be seen that the three- and four-core cable voltage drop per amp per metre (mV/A/m) is times the two-core (mV/A/m). The converts to three-phase; the division by two is necessary as there is assumed to be no voltage drop in the neutral."  Hence, where the 0.866 figure comes from which i have read about.


However, the Amtech handbooks simply divides the Z values in the table by root 3, without the division of 2 to get a single phase volt drop. Which makes more sense to me, as there is likely to be current in the neutral of the three phase system that needs to be accounted for. 


Example:


4 Core 3P+N 25mm2 cable supplying a distribution board 10m long, L1 110A, L2 80A, L3, 80A:

From table 4D4B z = 1.5


VD 1L1 = ( mv x L x I)/ 1000

VD = ((1.5/1.732) x 10 x 110)/1000

VD = 0.96v for the drop of the loads between L1 and Neutral


Is this correct, doesnt seem quite right to me.


Thanks



Parents
  • The saving grace is that it almost never matters - really big 3 phase loads tend to be well balanced, and the drops to unbalanced loads tend to be dominted by the voltage drop in the final leg of the circuit that is single phase.

    It is further confused by folk trying to look at the drop in the line to line voltage (400V - 5% or whatever) without realising that would be the exact same load that pulls the L-N voltage by the same % (i this case 230V -5%)

    You do need to know the waveforms ,  not just the currents,  before you can comment on the neutral cancellation or not. Clearly a circuit with lots of rectifiers feeding capacitors or charging batteries only draws current at the crest of the wave, so the peak current is a lot higher than 1.4 times the RMS, and 3 phases loaded with  a current  waveform like that cannot cancel, even if all are equal, as the 3 pulses of neutral current simply occur at different times.

    3 loads  numerically with a similar power factor but caused by a current to voltage phase shift from a very inductive load, would cancel just as well as the current from 3 resistors.

    To calculate accurately more info than you have available is needed, so the rules of thumb for various cases are pressed into service.

Reply
  • The saving grace is that it almost never matters - really big 3 phase loads tend to be well balanced, and the drops to unbalanced loads tend to be dominted by the voltage drop in the final leg of the circuit that is single phase.

    It is further confused by folk trying to look at the drop in the line to line voltage (400V - 5% or whatever) without realising that would be the exact same load that pulls the L-N voltage by the same % (i this case 230V -5%)

    You do need to know the waveforms ,  not just the currents,  before you can comment on the neutral cancellation or not. Clearly a circuit with lots of rectifiers feeding capacitors or charging batteries only draws current at the crest of the wave, so the peak current is a lot higher than 1.4 times the RMS, and 3 phases loaded with  a current  waveform like that cannot cancel, even if all are equal, as the 3 pulses of neutral current simply occur at different times.

    3 loads  numerically with a similar power factor but caused by a current to voltage phase shift from a very inductive load, would cancel just as well as the current from 3 resistors.

    To calculate accurately more info than you have available is needed, so the rules of thumb for various cases are pressed into service.

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