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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

0 gkenyon over 5 years ago

If the DB is balanced, as assumed in the EIDG and the NICEIC calculations, the % volt-drop TO THE DB is the same for three-phase and single-phase, as the neutral current is approximately zero - therefore the Neutral doesn't come into it.

For three-phase volt-drop in final circuits, the Neutral current still doesn't come into it, so just add the final circuit volt-drop to the DB volt-drop.

For single-phase circuits in a balanced system, all you need to do is add % volt-drop for the final circuit, to the % volt-drop at the DB.

When the system is highly imbalanced, this assumption for single-phase final circuits doesn't hold true, as there is now a Neutral current. You will need to follow one of the methods I gave in the earlier posts to examine the impact of this.

Either "take the biggest and overestimate" (second but easier method), or do the complete phasor calculation - but the maths is hard.
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0 gkenyon over 5 years ago

Or, put another way.

(a) You can't use examples that assume a balanced three-phase DB to try and understand a method that encapsulates single-phase and three-phase final circuits, when the DB is very imbalanced. This does not make the method for calculating volt-drop assuming a suitably-well balanced three-phase DB wrong - and even a fair few % imbalance could be ignored.

(b) You should avoid getting three-phase volt-drop (Line to Line) confused with single-phase volt-drop (Line to Neutral). However, this does not invalidate the method of calculating a % volt-drop to the DB three-phase being used in a balanced system ... BUT to convert the actual volt-drop from three-phase actual (calculated at 400 V line to line) to single-phase volt-drop (L-N), divide by √2 (or multiply by 0.707). If you calculate on percentages, as per EIDG, for a roughly balanced three-phase DB, for single-phase circuits you can simply add the % volt-drop to the DB, to the % volt-drop of the final circuit, to get the total % volt-drop. BUT this "add the %" method cannot simply be used in a highly imbalanced system.
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Votlage Drop for Unbalanced three phase

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