Correction factor Ct

[JohnW2]

It's an old Amicus book which also quotes 7.41 @ 20°C.

That (for one conductor) is obviously identical to the 14.82 (for two conductors) I quoted from the IET book.

However, as with the BGB the figures don't tally. It states 18.1 for 1mm² and 1.83 for 10mm². 18.1 / 2.5 = 7.24 x 2 x 1.2 = 17.376. Even a bit less with your 0 00393.

Indeed - your figure differs from mine because you are using a temperature correction factor of 1.2, whereas the true correction factor (assuming a coefficient of 0.00393 per degree) should be:

1 + (0.00393*(70-20)) = 1.1965.

Kind Regards, John

 

[JohnW2]

Indeed - your figure differs from mine because you are using a temperature correction factor of 1.2, whereas the true correction factor (assuming a coefficient of 0.00393 per degree) should be:
1 + (0.00393*(70-20)) = 1.1965

I'm actually just noticed a footnote to the 0.00393 figure - which says that 'a figure of 0.004 is used for temperature correction tables in many publications' - and that would result in your 1.2 figure.

Kind Regards, John

 

[EFLImpudence]

Yes but that's only a very small difference.

Much larger is the difference between 7.41 and 7.24 (18.1/2.5).

 

[ericmark]

Once started I wanted to get it right however one thing doing the calculations has shown me is how hard it would be to show with any certainty an installation was non compliant.

Most meters will give a three figure result but does not really matter if current or impedance do it three times and you will get three different readings.

As you have pointed out temperature will make a huge difference so the installation would need to be cold before measuring as there is no real way of measuring the temperature of the copper throughout the circuit with items like a final ring.

If the calculation showed 25 volts drop under load conditions then OK so well out one could comment on it with a EICR but at 15 volts drop it would be very hard to prove it was really that far out.

When a machine has failed in some way then measuring the short circuit current or loop impedance would show if volt drop was a likely cause.

If I think back I think twice there has been a machine failure due to volt drop. Once on a caravan site with main hum on radios and TV's plus I transmitted a really bad signal although not out of frequency. It was a DNO supply problem and since a temporary licence only no one was going to spend the money required to rectify. The second was a shrink wrap machine which was running too cool due to volt drop and there was really no need to calculate in either situation a volt meter was good enough.

Non HF florescent lamps are the main problem with volt drop and so much depends on location. Worst I found was building of Sizewell 'B' right next to the 'A' station when 'A' was on line 256 volt per phase common and when off line 230 volts but in the main supply voltage is well within limits and long cable runs may be giving a volt drop over the permitted but still not a problem.

However often we think volt drop may be causing a problem, so to be able to measure with a loop impedance meter and use HTML on a phone to quickly work out likely volt drop will help in identifying likely causes of a problem.

 

[JohnW2]

Yes but that's only a very small difference. Much larger is the difference between 7.41 and 7.24 (18.1/2.5).

True, but the difference would be smaller if you worked from their 10mm² figure - 1.83*10/2.5 = 7.32. If you took into account the fact that "1.83" could be the rounded version of anything up to virtually 1.835, you could then get 7.34.

As I said, even the differences you're now talking about are pretty academic.

Kind Regards, John

 

[ericmark]

My thanks to both JohnW2 and EFLImpudence without whom I would have never completed this project.

On the base program I have agnolaged all the help given by JohnW2 and EFLImpudence OK there were two others who gave one response but it was you two who did all the work in identifying my errors.

I think the project is now complete! I have included the java script on the main page so people can copy the whole page and use it to help them decide what are acceptable limits.

It has been great fun building the 4 pages and it has taught me more about java script I had always though 2 + 2 = 4 now I know 2 + 2 = 22 and eval 2 + eval 2 = 4. How could I get such a simple sum wrong.

With all this technical stuff I completely missed a / instead of a * and without JohnW2 I may have known there was an error but hadn't a clue what it was.

I consider this a team effort and can't praise JohnW2 and EFLImpudence enough for their help.

Eric GW7MGW

 

[EFLImpudence]

Thank you Eric.

Shouldn't you delete your personal links?

 

[ericmark]

To delete personal links means anyone finding an error can't report it. So I think personal links should remain as they are.

 

[JohnW2]

Thank you Eric. Shouldn't you delete your personal links?

Yes, well done, eric, and you're welcome as regards whatever assistance I've been able to offer.

EFLI, as regards links, are you perhaps partially referring to the hyperlinks to your and my DIYnot profile pages. It certainly doesn't matter to me - although I don't think it will be of my interest to anyone, either!

Eric's website is so 'exposed' (publicly accessible, and findable by search engines), anyway (even his call sign allows him, and his address etc.) to be identified (which is one reason why my call sign remains a 'secret' here!) - so I suppose it doesn't make much difference.

Kind Regards, John

 

[JohnW2]

It has been great fun building the 4 pages and it has taught me more about java script I had always though 2 + 2 = 4 now I know 2 + 2 = 22 and eval 2 + eval 2 = 4. How could I get such a simple sum wrong.

Yes, I guess the clue in in the name. Java Script was initially conceived as a scripting language, dealing primarily with character strings, and therefore didn't understand much about numbers and mathematics! Mind you, conversely, some of the more numerically-orientated languages aren't too hot at character/string operations, so maybe one can't win! However, some languages are fairly 'intelligent' and can automatically convert between numeric and character thinking depending upon what they are being asked to do. Look at this:

data _null_ ;
a = '2' ;
b = '3' ;
c = a || b ;
d = a + b ;
put 'c=' c ;
put 'd=' d ;
run ;

c=23

NOTE: Character values have been converted to numeric values

d=5

Kind Regards, John

 

[EFLImpudence]

EFLI, as regards links, are you perhaps partially referring to the hyperlinks to your and my DIYnot profile pages.

No, not that.

There are links to Eric's personal property along the top of the page on the calculator.

 

[JohnW2]

No, not that. There are links to Eric's personal property along the top of the page on the calculator.

Fair enough - but, as I said, his home page (with all those same links to other pages etc.) is publicly accessible (and easily Googled), anyway, so I presume he's not too concerned.

Kind Regards, John

 

[EFLImpudence]

Oh, I see. I've not been there. ok.

 

[JohnW2]

Oh, I see. I've not been there. ok.

If you click on the 'Index' link at the top of the calculator page, you'll get there!

Kind Regards, John

 

[JohnW2]

As I suspected, a rigorous mathematical examination of a circuit with a distributed load is far from trivial, but I’ve tackled ‘stage 1’.

With a bit of effort I have now satisfied myself that if one has a 32A circuit (ring or radial) with a 20A load at the furthest point, and the other 12A evenly distributed along the length of the circuit (equally in the two arms, in the case of a ring), and IF one assumes that the cable’s resistivity (mΩ/A/m) is the same throughout the cable, then the VD at the furthest point will be the same as one would calculate assuming a 26A load at the furthest point.

However, in the real world, the answer one gets by that calculation may not (I haven’t satisfied myself about this yet) be precisely correct, since current flow (hence temperature, hence conductor resistivity) will be different at every point in the cable. Not only that, but the current (hence corresponding temperature and resistivity) will rise to the ‘full’ 32A (or 2 x 16A for a ring), rather than the ‘quasi design current’ of 26A, very close to the origin of the circuit. Half of the cable length will be carrying a current greater than the 26A ‘quasi design current’, so half of the cable will have a resistivity (hence contribution to total VD) higher than that one would expect if 26A were flowing through the whole length of the cable, but the other half will be carrying less than 26A, hence lower temp/resistivity and less contribution to total VD than would be expected with a current of 26A flowing through the entire length of the cable. It remains for me to try to work out whether these two things exactly ‘cancel’....

... I suspect they may not cancel, since current varies in a linear fashion along the cable, but power dissipated, hence temperature, is presumably related to I² - so that the increased heating (hence increased VD contribution) in the more upstream parts of the circuit is probably less than the decreased heating (hence decreased VD contribution) in the more upstream parts. Watch this space.

Kind Regards, John