Correction factor Ct

[ericmark]

I understand your thoughts, but consider does it matter? One of the things my calculation pointed out to me was the loop impedance or short circuit current measurement will vary each time one measures, so in real terms the calculated result is an approximation and really only an indicator.

Yes it gives a warning that one is close to the limit or over the limit, but it's not good enough to tell anyone they are non compliant.

One would need to measure when nothing has been used for hours to ensure all the cable was cold. In real terms few cables have the same installation method throughout their run and flit between methods and one sees how the results change.

So where there is a fault likely caused by volt drop then the calculation will likely confirm or reject ones suspicions but at best it's only an indication.

 

[JohnW2]

I understand your thoughts, but consider does it matter?

If you're talking about the VD calculation (which I was), I don't think it matters in the slightest. As I discuss below, it's delivered voltage which matters, not VD. The 'guidance limits' for voltage drop are clearly totally arbitrary (and, generally, very conservative, given the permitted supply variation), and I therefore think that the most rough/approximate of VD estimates (or even a 'guestimate') is, in practice, more than adequate (and probably pretty meaningless).

One of the things my calculation pointed out to me was the loop impedance or short circuit current measurement will vary each time one measures, so in real terms the calculated result is an approximation and really only an indicator.

That's obviously true, as anyone who has ever measured any of these things will know, but I'm not sure what it has got to do with what we've been discussing.

One would need to measure when nothing has been used for hours to ensure all the cable was cold.

I'm not quite sure what you're suggesting, since that would surely be a very bad thing to do. The measurements of VD, Zs or whatever you got would then be lower than they would probably ever be in normal service - so the circuits could be compliant in terms of your 'cold measurements' but non-compliant the moment someone started using the circuits. I think I may have missed your point!

So where there is a fault likely caused by volt drop then the calculation will likely confirm or reject ones suspicions but at best it's only an indication.

Maybe it's just my lack of experience, but I really am not sure what (or how common) these 'faults due to voltage drop' actually are. If one suspects that a piece of equipment may be malfuncting because of VD in the circuit supply it, one surely would measure the voltage at its terminals to see? As I keep saying, it's really voltage, not 'voltage drop' which matters - a drop of, say, 6% from a supply voltage of 216V has a whole different set of possible implications than does a 6% drop from a supply voltage of 253V!!

Kind Regards, John

 

[ericmark]

One would need to measure when nothing has been used for hours to ensure all the cable was cold.

I'm not quite sure what you're suggesting, since that would surely be a very bad thing to do. The measurements of VD, Zs or whatever you got would then be lower than they would probably ever be in normal service - so the circuits could be compliant in terms of your 'cold measurements' but non-compliant the moment someone started using the circuits. I think I may have missed your point!

Sorry did not make it plain my point is few circuits will have a single installation method so only a cold measurement would cater for the mixture.

So where there is a fault likely caused by volt drop then the calculation will likely confirm or reject ones suspicions but at best it's only an indication.

Maybe it's just my lack of experience, but I really am not sure what (or how common) these 'faults due to voltage drop' actually are. If one suspects that a piece of equipment may be malfuncting because of VD in the circuit supply it, one surely would measure the voltage at its terminals to see? As I keep saying, it's really voltage, not 'voltage drop' which matters - a drop of, say, 6% from a supply voltage of 216V has a whole different set of possible implications than does a 6% drop from a supply voltage of 253V!!

Kind Regards, John

I have had two cases where volt drop within the installation was a problem. One a caravan site and it was also the supply causing a problem as well so voltages of around 190 were common because the voltage fluctuates it requires two meters to be read at the same time one on supply and one at the point of use to determine it the local distribution network around the site is out of spec or use either Ze/Zs or short circuit current as these are relatively stable as the load varies so can be compared.

The second was in a Jeyes factory now closed with a plug in shrink wrap machine. It worked in one location OK, but had problems in the other location. Without moving the machine back to the location where it failed I wanted to check the volt drop. Here the loop impedance meter was the only way, other than fork lift and returning the machine to that location. Again I had warned the foreman before installing the cable I thought it was too thin so could have used one of the other version before installation to show the problem, but my problem was to prove to the foreman that the cable was the problem not the machine. However did not really need the correction of the mV/A/m without correction it proved my point better.

In the main the supply to a distribution point is with a much heaver cable than from the distribution point to point of use so a simple Zs reading on it's own is no good it needs to be Zs - Ze to show the local volt drop and as you say when equipment is already installed simple volt meter is the way to go. The calculation only helps when the equipment is not installed either having been sent for repair or still has not arrived.

What I have been saying is before making the calculator I felt potential volt drop should be measured on a EICR and suspect circuits identified. However since making the calculator I have changed my mind as it would be too dependent on meters and unless looking at double the permitted volt drop it would be hard to determine if measurement or calculation error or real volt drop problem.

 

[JohnW2]

Sorry did not make it plain my point is few circuits will have a single installation method so only a cold measurement would cater for the mixture.

I still don't fully understand your point, since, whatever the characteristics of the circuit (e.g. installation methods), you surely want/need to know what it Zs and VD etc will be when the cable is at operating temperature when carrying the maximum normal load, don't you?

Kind Regards, John

 

[ericmark]

Sorry did not make it plain my point is few circuits will have a single installation method so only a cold measurement would cater for the mixture.

I still don't fully understand your point, since, whatever the characteristics of the circuit (e.g. installation methods), you surely want/need to know what it Zs and VD etc will be when the cable is at operating temperature when carrying the maximum normal load, don't you?

Kind Regards, John

I am rather confused now I thought the correction factor Ct gave me cold figures for the mV/A/m? There are only two points to measure the Zs either at full operating temperature in which case the whole calculator is useless as if on full load volt meter will measure direct or no load in which case then the calculation works as it will predict the volt drop once load is applied.

So what it seems your saying is my calculator needs to be filed with the chocolate fire guard?

 

[JohnW2]

I am rather confused now I thought the correction factor Ct gave me cold figures for the mV/A/m?

Using that equation, it will give you a figure for mV/A/m based upon its estimate of the conductor temperature at the design current you specified. If you wanted a 'cold figure' for mV/A/m (i.e. conductor temperature = ambient temperature), I think you would have to enter the 'design current' as zero - as I explained earlier, if you have ambient temp at 30° and max conductor temp as 70°, it will then give you CT ~0.87, which when multiplied by the BS7671-tabulated 70° mV/A/m figure will give you the mV/A/m figure for 30°.

There are only two points to measure the Zs either at full operating temperature in which case the whole calculator is useless as if on full load volt meter will measure direct or no load in which case then the calculation works as it will predict the volt drop once load is applied....

I'm not totally sure how Zs comes into this - do you mean you use Zs and Ze to determine (Zs-Ze), and then use that, together with knowledge of the CSAs of live conductors and CPC, to work out R1+Rn? The trouble with that is that, although that (R1+Rn) is what would theoretically determine the VD at the conductor temperature at which you took the measurement (for a near-zero current), the moment any significant current flows the conductor temp will rise and therefore the (R1+Rn) will no longer be the same as what you estimated/'measured'.

...or no load in which case then the calculation works as it will predict the volt drop once load is applied.

I'm getting very confused as to whether we are calculating VD (from cable size, length, design current and resistivity {manifest as mV/A/m}) or in some way determining it by indirectmeasurement (e.g. of Zs and Ze, as above). Perhaps you can clarify?

So what it seems your saying is my calculator needs to be filed with the chocolate fire guard?

I very much doubt that will be necessary. However, I have to say that I've never fully understood exactly you wanted/hoped to achieve with it - and if I could understand that better, maybe I'd be able to make some further suggestions.

Kind Regards, John

 

[ericmark]

Where Zs is calculated
Zs = Ze + cable length x (corrected mV/A/m for earth + corrected mV/A/m for line)/2 It also is divided by 1000 to get to volts and by 2 twice when ring selected.

Where Zs is given the Ze - Zs divided by (half corrected mV/A/m for earth + half corrected mV/A/m for line).

 

[JohnW2]

Where Zs is calculated ...
Zs = Ze + cable length x (corrected mV/A/m for earth + corrected mV/A/m for line)/2 It also is divided by 1000 to get to volts and by 2 twice when ring selected.

OK. That will give you the Zs at whatever temperature you have corrected the mV/A/m figures for - but that leads one having to decide what temperature to use for the correction. [the "/2" obviously arises because you are starting with a mV/A/m figure which relates to 2 x line conductors].

Where Zs is given the Ze - Zs divided by (half corrected mV/A/m for earth + half corrected mV/A/m for line).

OK [you've used the 'half's this time, instead of the "/2"] - if you correct mV/A/m to the temperature at which the Zs measurement was undertaken that calculation should give you cable length. Is that what you want?

However, I'm still not completely sure of what you are trying to achieve. Extending what I wrote last night, I think (haven't checked it rigorously yet) you could get from a measured 'cold' Zs (no appreciable current having flowed for some time) and a measured Ze to the VD at your chosen Ib as follows:

1...Measure Ze and 'cold' Zs, hence determine (Zs-Ze)

2...Convert (Zs-Ze) to (R1+Rn) (at 'cold' ambient temperature) as follows:

calculate k = (uncorrected mV/A/m for line / uncorrected mV/A/m for CPC)
R1 (at ambient temp) = (Zs-Ze) * (k/(1+k))
(R1+Rn) (at ambient temp) = R1 * 2

[you could obviously combine all those steps into a single equation if you wanted]

3...To estimate the (R1+Rn) when current Ib is flowing, calculate:
(R1+Rn) {at ambient temp, from above} * (Ct for your Ib) / (Ct for Ib=0)

4...To estimate VD when Ib flowing, multiple the temperature-corrected figure from (3) by Ib.

Would that help you? I'll do some checking, hopefully to confirm that it's actually correct!

Kind Regards, John

 

[ericmark]

I will need to think about this. Had not thought of using k but it would mean starting from scratch. Some time in the future I may give it a go but other matters now pressing so that will be some time in the future I think this project is now complete and it does well enough for me.

The only point for me to consider is do I leave on web site or remove. The putting on web site allowed you to test results and find my many faults but is it good enough to leave for others to use or should I remove it?

 

[JohnW2]

I will need to think about this. Had not thought of using k but it would mean starting from scratch.

You don't have to use a 'k'. I merely did it that way to stop the equations looking horrendous, particularly in a forum post! You could combine all of my (2) into a single equation (without any intermediary 'k') if you wanted. Indeed, you could combine all four steps into one equation if you really wanted (I'll do it for you, if you want) - but I personally find it much easier to read/follow when it is broken up into 'bite sized pieces'!

Some time in the future I may give it a go but other matters now pressing so that will be some time in the future I think this project is now complete and it does well enough for me.

Fair enough. As I've said, I might well be able to help you more if I really understood what you are trying to achieve.

The only point for me to consider is do I leave on web site or remove. The putting on web site allowed you to test results and find my many faults but is it good enough to leave for others to use or should I remove it?

As far as I can make out, it probably more-or-less does 'what it says on the tin'. The main question is whether it is actually calculating what you (and others) want/need to be calculated - and I can't answer that.

Whatever, I remain of the view that (if that's what it's about) very precise calculation/estimation of VD is probably more of academic than practical interest.

Kind Regards, John