Volt Drop

[Lectrician]

In case I am going barking mad.

Take a circuit rated at 70amps per phase, three phase and neutral.

5% volt drop max (of 230v as neutral required).

What volt drop figure do you get for a circuit running 150m run in 25mm?

I calculate it to be 15.75 volt, using 1.5mV/a/m

1.5 x 70 x 150
-----------------
1000

This is what I would expect.

With my very old amtech calc software, this agrees.

With the new amtech calc software from last year, I get this:

Untitled

• Lectrician
• 8 May 2013
• 1

 

[securespark]

Roughly half. I'm lost!

 

[JohnW2]

In case I am going barking mad. Take a circuit rated at 70amps per phase, three phase and neutral. 5% volt drop max (of 230v as neutral required).
What volt drop figure do you get for a circuit running 150m run in 25mm?
I calculate it to be 15.75 volt, using 1.5mV/a/m
1.5 x 70 x 150
-----------------
1000
This is what I would expect. With my very old amtech calc software, this agrees. With the new amtech calc software from last year, I get this: [8.45V]

Does the lower figure from your new software not perhaps assume a perfectly balanced 3-phase situation?

What you have calculated manually (15.75V) is the voltage drop you will get if 70A flowed through both neutral and phase conductors. If the circuit were perfectly balanced, the net current through the neutral conductor (hence also VD due to neutral conductor) would be zero, hence VD would presumably be half of your calculated value.

...or have I got this totally wrong?!

Kind Regards, John

 

[ricicle]

In the absence of any PF data you would use the 1.65 mV/A/m (impedance Z) figure

(1.65x70x150) / 1000 = 17.325v

It doesn't matter about the neutral, as it is a three phase supply you use 400v so 5% is 20v so you are just below that figure.
This assumes there are no starting current/voltage issues !

 

[JohnW2]

In the absence of any PF data you would use the 1.65 mV/A/m (impedance Z) figure
(1.65x70x150) / 1000 = 17.325v

As per what I wrote above, whether one uses 1.65 mV/A/m or 1.5 mV/A/m figure, doesn't theat figure relate to the situation in which 70A flows in two conductors of length 'm'? (i.e. if it were single phase, 8.6625V drop in L and 8.6625V drop in N, making your 17.325V total drop)

Kind Regards, John

 

[Lectrician]

Its three phase AND neutral, so volt drop must be treated as 230v.

The software can be set to TP + N or just TP, and was set to TP + N, same results if I tried to go with single phase too.

PF on the software set to 1, so used the 1.5 figure in the regs.

This is a theoretical question, as I have had a few issues with the software not calculating as it should when double checking with the regs.

 

[JohnW2]

The software can be set to TP + N or just TP, and was set to TP + N, same results if I tried to go with single phase too.

If it's giving the same ('low') result for single-phase (two conductor L+N), then that's certainly wrong - since your manual calculation was certainly correct for that.

Kind Regards, John

 

[ricicle]

It doesn't matter about being 3 phase AND neutral. You use 400v!

 

[Lectrician]

I don't agree with that Riccle. If something needs the neutral for 230v, you must do your calc on that. When calculating a submain for example, you will have single phase loads.

 

[ricicle]

Are you saying that there could be a situation where there is 70A on one phase and zero on the other two then? I never use design software. But look at the two tables for VD in Bs7671.

 

[JohnW2]

It doesn't matter about being 3 phase AND neutral. You use 400v!

I'm not sure about interpretation of the regs for a 3-phase load, but that would make little sense if one connected a single-phase load between one of the plases and neutral (which is, after all, what's happening in any single-phase installation).

In any event, there surely is no difference between a 5% drop in the 230V between phases and neutral and a 5% drop in the 400V between-phases?

Kind Regards, John

 

[Lectrician]

There is every chance, especially at an early commissioning stage.

I have always calculated circuits that require a N at 230v and not 400v, perhaps I could of saved some copper

 

[ricicle]

I added to my post above to include my reasoning!

 

[plugwash]

IMO it's really a case of "risk assesment". To properly work out the volt drop in the neutral you need to know both the likely maximum load and the likely degree of phase imbalance in that load.

 

[JohnW2]

IMO it's really a case of "risk assesment". To properly work out the volt drop in the neutral you need to know both the likely maximum load and the likely degree of phase imbalance in that load.

Indeed - and, as I said right at the start, if the software in question were assuming a perfectly balanced load, then the current (hence VD) in the neutral would be zero, and the only VD to take into account would be that in each of the phase conductors.

With PF=1, the VD in a single 25mm² conductor is 0.75 mV/A/m. Hence, with a perfectly balanced load drawing 70A down each of the three 150m long phase conductors, the VD along each of those conductors would be 7.875V (i.e. half of Lectrician's manual calculation). The voltage at the end of each phase conductor, relative to neutral, would hence have dropped by about 3.42% (7.875/230*100). If you prefer, you can view that at the between-phases voltage at the load having dropped by about 13.64V (7.875*√3) - again, a VD of about 3.42% (13.64/400*100).

At the other extreme, if there is 70A down one phase and zero down the other two (hence 70A down neutral), then Lectrician's calculation would be correct. The VD along the pair of used conductors (one phase + neutral) would then be 15.75V, a drop at the end of that one phase conductor of about 6.84% (phase-neutral).

So, as you say, the VD will depend upon the balance of the load, as well as its magnitude, and I guess one has to design for the worst likely case of phase imbalance.

Kind Regards, John