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Votlage Drop for Unbalanced three phase

fiftyhertz
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



  • 0
    Neither of the methods you have discussed is 100 % accurate for an unbalanced system. In some installations, knowledge of the power factor of the loads on the phases (single-phase and three-phase) might also skew the answer.


    The reason for this, is the actual voltage on the Neutral "shifts" in phase due to the imbalanced load.
  • 0

    gkenyon:

    Neither of the methods you have discussed is 100 % accurate for an unbalanced system. In some installations, knowledge of the power factor of the loads on the phases (single-phase and three-phase) might also skew the answer.


    The reason for this, is the actual voltage on the Neutral "shifts" in phase due to the imbalanced load.




    Graham, thanks firstly for the response. I understand PF would affect the drop, especially the larger cables with x and r values. If its not too much trouble could you advise on the best method of calculating the above (or similiar) potentially just with z and PF unity for initial clarity. I hate just putting values into Amtech, and not having a firm understanding of the calculation

  • 0
    The only way to do it accurately is to know each line current and its associated power factor.


    The process in reality uses the following steps:


    (a) Calculate the volt-drop to the DB due to the line current using the single-phase volt-drop / 2 - you will need to keep the phasor value of the volt-drop.


    (b) Calculate the Neutral current using the vector sum of IL1+IL2+IL3, taking into account the power factor of each phase


    (c) Calculate the Line to Line voltages using the vector sums (the largest of these is the three-phase volt-drop).


    (d) Calculate the worst case of each VLx-Vn and this is the worst-case single-phase voltage drop at the DB


    The downstream three-phase and single-phase voltage drops can then be calculated by adding those using the standard method (balanced three-phase, and single-phase) to the worst-case volt-drop at the DB, for each circuit.


    This I guess is why we try to balance each db as far as possible.



    The methods in BS 7671 are based on certain assumptions, and cannot be used for every case I guess.
  • 0
    So, in terms of the unbalanced system, what's going on is that the line currents don't cancel in the Neutral.


    What happens then, is that, at the DB, the Neutral moves not just in the magnitude of the voltage, but the phase (as simply the line currents are around 120 degrees apart). So, knowing the power factors of the phase makes it far more accurate.
  • 0
    Now, if all that's too much, there is an easier way if you only want the worst-case volt-drop you could imagine, although this could potentially be a huge over-estimate.


    (a) Take the largest line current for each phase you would envisage at the DB. Take the biggest one.


    (b) Calculate the single-phase voltage drop. This will give you a larger result than the above calculation - an over-estimate, but very easy calculation.


    (c) Calculate the three-phase voltage drop using this largest line current as the line current and assume it's balanced. This will give you the three-phase (line to line) voltage drop.


    (d) Calculate the final circuit voltage drops from each of the circuits downstream, added to the relevant single-phase and three-phase values at the DB calculated in (b) and (c) respectively.
  • 0

    gkenyon:

    Now, if all that's too much, there is an easier way if you only want the worst-case volt-drop you could imagine, although this could potentially be a huge over-estimate.


    (a) Take the largest line current for each phase you would envisage at the DB. Take the biggest one.


    (b) Calculate the single-phase voltage drop. This will give you a larger result than the above calculation - an over-estimate, but very easy calculation.


    (c) Calculate the three-phase voltage drop using this largest line current as the line current and assume it's balanced. This will give you the three-phase (line to line) voltage drop.


    (d) Calculate the final circuit voltage drops from each of the circuits downstream, added to the relevant single-phase and three-phase values at the DB calculated in (b) and (c) respectively.




    Is this saying then that the total VD on, for example, L1 on an unbalanced 3ph distr cct with single ph final cct, is the single ph VD in the distr cct + the 3ph VD in the distr cct + the single ph VD of the final cct, all on L1?


    Is there an example of calculating VD for unbalanced three ph system in the EIDG, i can't seem to see it? 


    Also, have they used the correct methodology in this article then? They seem to be calculating VD in the 3ph distribution cct for balanced and unbalanced systems, but don't make it explicit, yet use the 3ph mV for a balanced system. Sorry if I've misunderstood it.


    professional-electrician.com/.../


    F

     

  • 0
    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.

  • 0

    Farmboy:



    Is this saying then that the total VD on, for example, L1 on an unbalanced 3ph distr cct with single ph final cct, is the single ph VD in the distr cct + the 3ph VD in the distr cct + the single ph VD of the final cct, all on L1?


    Is there an example of calculating VD for unbalanced three ph system in the EIDG, i can't seem to see it? 


    Also, have they used the correct methodology in this article then? They seem to be calculating VD in the 3ph distribution cct for balanced and unbalanced systems, but don't make it explicit, yet use the 3ph mV for a balanced system. Sorry if I've misunderstood it.


    professional-electrician.com/.../


    F

     


     




    The three-phase volt drop and single-phase volt drop are two separate things.


    The three-phase volt-drop is the volt-drop between Line conductors.


    The single-phase volt-drop is the volt-drop between Line and Neutral.


    Hence, the Neutral is never taken into account for a three-phase volt-drop, and always taken into account for the single-phase.

  • 0
    Sorry ... got diverted onto something else

     

    Is there an example of calculating VD for unbalanced three ph system in the EIDG, i can't seem to see it? 





    No, page 65 para 2 explains why only a balanced option is used.



    Also, have they used the correct methodology in this article then? They seem to be calculating VD in the 3ph distribution cct for balanced and unbalanced systems, but don't make it explicit, yet use the 3ph mV for a balanced system. Sorry if I've misunderstood it.


    professional-electrician.com/.../





    This follows the examples given in the EIDG - which assumes three-phase volt drop to the find DB and single-phase for the single-phase final circuit.


    However, as I've said, if the system is so badly imbalanced (which may or may not cause other problems in the installation), then you can't really follow the broad guidance in the EIDG.




  • 0

    gkenyon:

    Sorry ... got diverted onto something else

     




    Is there an example of calculating VD for unbalanced three ph system in the EIDG, i can't seem to see it? 





    No, page 65 para 2 explains why only a balanced option is used.



    Also, have they used the correct methodology in this article then? They seem to be calculating VD in the 3ph distribution cct for balanced and unbalanced systems, but don't make it explicit, yet use the 3ph mV for a balanced system. Sorry if I've misunderstood it.


    professional-electrician.com/.../





    This follows the examples given in the EIDG - which assumes three-phase volt drop to the find DB and single-phase for the single-phase final circuit.


    However, as I've said, if the system is so badly imbalanced (which may or may not cause other problems in the installation), then you can't really follow the broad guidance in the EIDG.




     

     




    Graham, firstly there is some excellent information in the above posts, however the one thing i dont quite understand is that previously you mentioned that three phase and single phase volt drop were two diffirent enterties and therefore for a single phase volt drop at a final circuit we should not mix and match as the example in proffesional electrician. As anyhting in the neutral from the final circuit imbalance will also be in the neutral of the submain. I understand that if the final circuit was three phase circuit such as a motor there will be no drop in the neutral so we can use the three phase volt drop in the submain for that calc.


    In addition, for the EIDG page 65 states that it assumes the fourth core should not be loaded if balanced. But if unbalanced and calcultaing three phase should we not calculate the three phase volt drop and the add the drop in the neutral if it unbalanced? Would this be correct


    Thanks again

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