Volt Drop

[EFLImpudence]

As the VD figures in the tables are the same for three and four core cables, is it that the neutral is considered as if the current were the same as the line - just in case?

 

[JohnW2]

As the VD figures in the tables are the same for three and four core cables, is it that the neutral is considered as if the current were the same as the line - just in case?

I think I would have guessed the opposite! If they're quoting the same VD for 4-core as for 3-core, I would have thought that would mean that they were assuming that the VD due to the neutral (hence the neutral current) was zero (i.e. perfectly balanced phases).

Kind Regards, John

 

[Lectrician]

From the regs:

For a given run, to calculate the voltage drop (in mV) the tabulated value of voltage drop per ampere per metre for the cable concerned has to be multiplied by the length of the run in metres and by the current the cable is intended to carry, namely, the design current of the circuit (lb) in amperes. For three-phase circuits the tabulated mV/A/m values relate to the line voltage and balanced conditions have been assumed.

 

[JohnW2]

From the regs:

... For three-phase circuits the tabulated mV/A/m values relate to the line voltage and balanced conditions have been assumed.

Ah, it's reassuring to see that I occasionally get something right

Kind Regards, John

 

[EFLImpudence]

As the VD figures in the tables are the same for three and four core cables, is it that the neutral is considered as if the current were the same as the line - just in case?

I think I would have guessed the opposite! If they're quoting the same VD for 4-core as for 3-core, I would have thought that would mean that they were assuming that the VD due to the neutral (hence the neutral current) was zero (i.e. perfectly balanced phases).

Then why are the values so similar to two-core single phase?

 

[JohnW2]

I think I would have guessed the opposite! If they're quoting the same VD for 4-core as for 3-core, I would have thought that would mean that they were assuming that the VD due to the neutral (hence the neutral current) was zero (i.e. perfectly balanced phases).

Then why are the values so similar to two-core single phase?

One of my uncertainties is exactly what they mean by 'voltage drop' in the 3-phase situation - I'll have a read of the text to see if it clarifies.

IF, as I rather suspect, the voltage-drop they are talking about is the drop in the between-phases voltage (nominally 400V), then I would expect that VD, (for a balanced 3-phase load) to be pretty close to that for the 2-core single-phase situation - more specifically, I would expect it to be about 86.6% of the VD for a 2-core single-phase situation (how does that compare with the Tables - I don't have them to hand?).

My reasoning: Consider a cable of length L with a voltage drop of D mV/A/m per conductor (e.g. 0.75 mV/A/m for 25mm²). For a 2-core single-phase situation, with current I flowing (in both L and N), the voltage drop will be 2*D*I*L mV. For an equally balanced 3-phase situation with the same conductor, with current I in each phase core (and zero current in neutral) and each of length L, the voltage drop along each core will be D*I*L mV. That translates to a drop in the between-phase voltage (at load end) of D*I*L* √3 mV. The ratio of the two is therefore:
(D*I*L*√3) / (2*D*I*L)
= (√3 √ / 2) = 0.866 (or 86.6% if you prefer)

Does that make sense, and is it close to the figures in the Tables?

Kind Regards, John

 

[EFLImpudence]

I would expect it to be about 86.6% of the VD for a 2-core single-phase situation (how does that compare with the Tables - I don't have them to hand?).

Yes, thereabouts.

My reasoning: Consider a cable of length L with a voltage drop of D mV/A/m per conductor (e.g. 0.75 mV/A/m for 25mm²). For a 2-core single-phase situation, with current I flowing (in both L and N), the voltage drop will be 2*D*I*L mV.

Aren't the figures in the VD tables (in BGB) already doubled, i.e. for both conductors.

Hence my query as to why one-phase and three-phase are similar (86%) and not roughly half (43%)[/quote]

 

[JohnW2]

I would expect it to be about 86.6% of the VD for a 2-core single-phase situation (how does that compare with the Tables - I don't have them to hand?).

Yes, thereabouts.

That sounds promising ... maybe I'm on the right track, then.

My reasoning: Consider a cable of length L with a voltage drop of D mV/A/m per conductor (e.g. 0.75 mV/A/m for 25mm²). For a 2-core single-phase situation, with current I flowing (in both L and N), the voltage drop will be 2*D*I*L mV.

Aren't the figures in the VD tables (in BGB) already doubled, i.e. for both conductors. ... Hence my query as to why one-phase and three-phase are similar (86%) and not roughly half (43%)

Yes, the figures (for 2-core) in the table are already doubled - but, as you've quoted me as saying, my 'D' was for a single conductor (i.e. half of the ('doubled') figures in the BGB Tables - e.g. the example I quoted: D=0.75 for 25mm², as compared with 1.5 in the BGB table).

Does that clarify?

Kind Regards, John

 

[EFLImpudence]

Yes, the figures (for 2-core) in the table are already doubled - but, as you've quoted me as saying, my 'D' was for a single conductor (i.e. half of the ('doubled') figures in the BGB Tables - e.g. the example I quoted: D=0.75 for 25mm², as compared with 1.5 in the BGB table).

Does that clarify?

Not entirely.
As the figures for three phase are 86% of single phase are they not also including the neutral?
If not, shouldn't they be 43% of the single phase?


Is lectrician's software halving the figures for three phase thus discounting the neutral half?

 

[JohnW2]

Yes, the figures (for 2-core) in the table are already doubled - but, as you've quoted me as saying, my 'D' was for a single conductor (i.e. half of the ('doubled') figures in the BGB Tables - e.g. the example I quoted: D=0.75 for 25mm², as compared with 1.5 in the BGB table). Does that clarify?

Not entirely. As the figures for three phase are 86% of single phase are they not also including the neutral? If not, shouldn't they be 43% of the single phase?

No. I wish I were with you and a blackboard - it would be so much easier

Lets do it with some easy figures. 10 metres of 25mm² and a current of 10A. The BGB overall (L+N) VD figure for 2-core single phase is 1.5 mV/A/m, so the total VD is 1.5*10*10 = 150 mV - i.e.a VD of 75 mV in the L conductor and 75 mV in the N one.

Now move to the 3-phase situation. Still 10m, with 10A in each of the three phase conductors, and zero in the neutral. The VD along each of the phase conductors will be the same as along one of the conductors in the single-phase case, namely 75 mV. So we have a drop of 75 mV at the ends of each of the three phase conductors. That corresponds to a drop in the between-phases voltage of (75*√3) = 129.9 mV.

We therefore have a VD (the definition of which is obvious) in the single-phase situation of 150 mV, and a VD (in terms of the between-phases voltage, if that's what they're talking about) of 129.9 mV in the 3-phase situation. 129.9 is about 86.6% of 150.

Am I being any clearer?

Kind Regards, John

 

[EFLImpudence]

Ah. penny's dropped.

86.6 = (√3)/2

Thanks.

 

[JohnW2]

Ah. penny's dropped. 86.6 = (√3)/2 Thanks.

You got it! I guess you did not notice, about 3 posts back, when I wrote (with a typo of a suiperfluous "√",which may have confused you!):

(D*I*L*√3) / (2*D*I*L)
= (√3 √ / 2) = 0.866 (or 86.6% if you prefer)
should have been
= (√3 / 2) = 0.866

However, none of this explains exactly what Lectrician's software is up to. His manual calculation, as appropriate for a 2-core single-phase set up (and using 1.5 mV/A/m), correctly gave a VD of 15.75V - i.e. 7.875V drop in each of L and N in single-phase 70A. For the balanced 3-phase situation (70A in each phase, zero in neutral), the VD of each of the phases would therefore be 7.875V and the VD in terms of the between-phases voltage would therefore be (7.875V * √3) = 13.64V. His software gave a VD of 8.45V, which is neither of those figures.

Kind Regards, John

 

[EFLImpudence]

One thing I have noticed in the results picture in the original post is -

that while it doesn't state the length of the cable or VD figure (you'd think it would)

it states that the volt drop is 8.45V and the maximum length for volt drop is 204.1m

For 230V single phase, which was being used, this is the correct proportion - 8.45 / 150 x 204.1 = 11.5V.

The machine would appear to be using a VD figure of 0.805mV/A/m - 1.5 x 8.45 / 15.75

I don't know what this proves, except that it is the VD figure which is going wrong, but does it ring any bells?

 

[JohnW2]

One thing I have noticed in the results picture in the original post is - that while it doesn't state the length of the cable or VD figure (you'd think it would) it states that the volt drop is 8.45V and the maximum length for volt drop is 204.1m

It is a rather odd display/printout we've been shown - I suspect it is only part of the whole. The software clearly knows about a lot of things it hasn't displayed in what we've been shown, since it's somehow calculated Zs (hence it must know about the CPC and 'Ze'), disconnection time (hence it must know about the OPD) etc. etc. There are all sorts of aspects of what we've been shown that I don't fully understand. I wonder if Lectrician might perhaps be able to provide us with a link to a manual for this software?

For 230V single phase, which was being used, this is the correct proportion - 8.45 / 150 x 204.1 = 11.5V.

Yes, that's interesting - but are we really sure it was assuming single phase? - I'm not sure what a 'cable makeup' of "1 x 1 x 4c" means, although I suspect that '4c' probably means 4-core - hardly 'single phase'. The absence of anything under the 'Sep CPC size' field might suggest that it is assuming that one of the 4 cores is a CPC, hence no neutral! However, the 'Earth Fault Current" has clearly been calculated from the Zs on the assumption of 230V, so I don't know.

The machine would appear to be using a VD figure of 0.805mV/A/m - 1.5 x 8.45 / 15.75 ... I don't know what this proves, except that it is the VD figure which is going wrong, but does it ring any bells?

The nearest to a bell that rings is that 0.805 mV/A/m is close to being half of the 1.65 mV/A/m figure which ricicle suggested should be used when one knows nothing about PF. If it were 1.65 mV/A/m for single-phase L+N, that would be 0.825 mV/A/m for each of the two conductors. If the software was actually thinking of balanced 3-phase (70A in each phase, zero in neutral {or no neutral!}) then, for a 150m run, the VD along each phase conductor (i.e. the VD in terms of phase-neutral voltage) would be calculated as 0.825 x 70 x 150 / 1000 = 8.66V, not too far off the software’s 8.45V. If ricicle's figure had been 1.61, it would have been 'spot on'.

However, that’s probably all just co-incidence, and hence a red herring.

Of course, one of the difficulties we (at least, I) are having in this discussion is that we (at least, I!) don’t really know for sure what anyone (including the BGB) mean when they talk about VD of a 3-phase situation – are they talking about a drop in the phase-neutral voltage or the (larger, by a factor of √3 in the balanced situation) drop in the phase-phase voltage?

I’ll continue pondering.

Kind Regards, John

 

[Lectrician]

OK.

While I am not at the office, I can remote desktop into my PC up there, and use the software.

I have just let it do an online update, and I now get the below:

150m, 70amp, forcing 25mm.

Set to:

Single Phase + N: 16.79v (assumes 2 core SWA)

Three Phase: 8.45v (assumes 3 core SWA)

Three Phase + N: 8.45v (assumes 4 core SWA)

Previously, it would give a different result for TP and TPN. It now doesn't. Even if I load some older saved calculations, they are re-calculated with the new version, as the 'save' feature only appears to save the variables.

I have always sized three phase sub-mains as if they were single phase. I was always taught this, and can recall doing some calcs on site during an NIC inspection to determine volt drop too. Now not too sure what I will do when I need to calc another

I think the screen shot I put up was the wrong one, and was for three phase, as I was prattling between the three types trying to work out why things didn't tally. There are several tabs to go through to change the external Ze, the cable type, installation method, cables in parallel, ambient temp, OPD, PF etc. Too many to screen shot everything.