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
