Correction factor Ct

[ericmark]

Correction factor Ct = ((nv+tp)-(POWER(ca,2)*POWER(cg,2)-POWER(ib,2)/POWER(it,2))/(tp-at))/(nv+tp)
Where nv = Nominal volts (230)
tp = Max permitted operating temp (70)
ca = Rating factor for ambient temperature (1)
cg = Rating factor for grouping (1)
ib = Design current for circuit (26)
it = Tabulated current-carrying capacity of a cable (42)
at = Ambient temperature (30)

For a ring final I can break this down to (300-(1/0.38 )/40)/300 = 0.9999 which is not enough to vary the mA/V/m by enough to really affect the cable length by enough to get 106 meter for a ring final.

11.5/0.0179/26*4 = 98 meters the V/A/m needs to drop to 0.0167 before I can get 106 meters which I was told was the maximum length of cable in a ring final. Even with 25 amp as design current for circuit it still needs the mV/A/m to drop to 17.4 which just seems too much to happen purely due to a correction for temperature.

When I first ran the formula I missed off a bracket Ct = ((nv+tp)-(POWER(ca,2)*POWER(cg,2)-POWER(ib,2)/POWER(it,2))/tp-at)/(nv+tp) and I got that 106 meters which leads me to ask all those years ago (2008) when at the IET lecture had that lecturer made the same mistake? Or is the design current for circuit Ib only 24 amp not 26 amp?

I was told 20 amp centre of ring and rest even so consider it as 12/2 so total = 26 amp.

I tried to make a program to work it out for different values and couldn't get the expected figures so have been trying to get any faults corrected but have reached the stage when I am wondering is there really an error in program or did I get wrong information all those years ago?

 

[EFLImpudence]

I am sure John will be along with more detailed suggestions but I am wondering if a calculator can be used for Ring finals as they are a special case.

When entering Method C the length can increase to 102m.
Not sure why this should be so but could one of your formulae be doing this?


EDIT - no it doesn't because I had altered the Ib as well - but is that any help?

 

[JohnW2]

When I first ran the formula I missed off a bracket Ct = ((nv+tp)-(POWER(ca,2)*POWER(cg,2)-POWER(ib,2)/POWER(it,2))/tp-at)/(nv+tp) and I got that 106 meters which leads me to ask all those years ago (2008) when at the IET lecture had that lecturer made the same mistake? Or is the design current for circuit Ib only 24 amp not 26 amp? ... I was told 20 amp centre of ring and rest even so consider it as 12/2 so total = 26 amp.

FWIW (I've now found my OSG, which was temporarily lost!), the OSG says the same as you are suggesting, namely:

The circuit is assumed to have a load of 20A at the furthest point and the balance to the rating of the protective device evenly distributed. (For a 32A device, this equates to a load of 26A at the furthest point)

I tried to make a program to work it out for different values and couldn't get the expected figures so have been trying to get any faults corrected but have reached the stage when I am wondering is there really an error in program or did I get wrong information all those years ago?

I haven't got much time at the moment, but I'll have a look into this later and see if I can offer any ideas/suggestions. Off the top of my head (and without yet having attempted any calculations), I'm wondering whether one really can use this "equivalent to a 26A load at the furthest point" to undertake VD calculations when the load is distributed, such that different currents are flowing through different segments of the cables. It could be that the 'averaging' approach works, but I need to convince myself of that. What is certainly true of a distributed load, is that the temperature rise due to current flow (hence conductor temperature) will be different at different points in the cable, and I suspect that may well mess all this up - since I think the equation you are using probably assumes that the same current-heating effect will be experienced throughout the length of the cable cable - however, again, it's just possible that the 'averaging' technique copes with that, too (but I'm suspicious, since local heating will presumably be a function of I², not I). I'll try to work this out.

Kind Regards, John

 

[ericmark]

Thank you for info on the on site guide that at least removes on possibility. I have no idea where the lecturer got the 106 meter from and my failure to emulate this figure to me means either I have got the 106 meters wrong or the calculation wrong.

To get 106 meters the Correction factor Ct needs to be 0.97 and this in turn is down to the relationship between Design current for circuit Ib and Tabulated current-carrying capacity of a cable It.

Using Installation method C one can get it to 102 meters tried using 31 amp i.e. mineral insulated but still can't top the 102 meters. Ct = 0.9999345278920896 which is the lowest reading I could get and that was cheating as I did not swap 70°C to 90°C.

I was happy until it was pointed out as the temperature rises then the resistance increases so you can use less not more cable.

I am however not convinced that the cable max temperature should make a difference as a bit of 2.5 with 13A should get to same temperature be it insulated with thermal plastic or thermal setting insulation so can't see why it matters what the max temperature can rise to if it does not ever reach that temperature?

 

[JohnW2]

I am however not convinced that the cable max temperature should make a difference as a bit of 2.5 with 13A should get to same temperature be it insulated with thermal plastic or thermal setting insulation so can't see why it matters what the max temperature can rise to if it does not ever reach that temperature?

It doesn't matter (well, very little - I suspect the different types of plastic have very similar thermal resistance/conductance properties), in terms of what the temperature will be for a given current. However, if you are using the tables in Appendix 4 of the regs for your VD figures (mV/A/m), they assume the maximum operating temperature for the type of cable (insulation) concerned.

The equation (for Ct) you are using appears to assume that the conductor temperature will be the 'maximum permitted operating temperature' if the 'maximum permitted current' is flowing, and attempts to work out what temperature will be reached (hence what adjustment to VD is required) if the current flowing is less than the permitted maximum for the cable (i.e. the 'CCC' per BS7671 tables).

Kind Regards, John

 

[JohnW2]

Correction factor Ct = ((nv+tp)-(POWER(ca,2)*POWER(cg,2)-POWER(ib,2)/POWER(it,2))/(tp-at))/(nv+tp)

Is that Excel pseudocode? Whatever, if I understand correctly what's going on, I think it probably contains an error. Without having tested it (I will later), I would suspect that it probably should be:

((nv+tp)-(POWER(ca,2)*POWER(cg,2)-POWER(ib,2)/POWER(it,2))*(tp-at))/(nv+tp)

or, easier on my eyes:

((nv+tp)-(ca^2*cg^2-ib^2/it^2)*(tp-at))/(nv+tp)

Frankly, since I personally tend to get nervous about being sure that calculations get done in the right order (particularly when jumping between programming languages), I would probably have thrown in some extra (actually unnecessary) brackets:

((nv+tp)-((ca^2*cg^2-((ib^2)/(it^2)))*(tp-at) ))/(nv+tp)

More later, particularly after I've tested the above.

Kind Regards, John

 

[EFLImpudence]

To get 106 meters the Correction factor Ct needs to be 0.97

18 x 0.97 = 17.46

Can I point out that the VD figures given in the BGB do not relate.

1mm² is 44, 10mm² is 4.4 so can we assume they are correct?

44 / 2.5 = 17.6


Perhaps the VD figures are just not accurate enough

 

[ericmark]

Thank you John your spot on missed it. It was times not divide and now rest seems to fall into place.

 

[JohnW2]

The equation (for Ct) you are using appears to assume that the conductor temperature will be the 'maximum permitted operating temperature' if the 'maximum permitted current' is flowing, and attempts to work out what temperature will be reached (hence what adjustment to VD is required) if the current flowing is less than the permitted maximum for the cable (i.e. the 'CCC' per BS7671 tables).

To illustrate what the equation is doing, if Ca=1, Cg=1, ta=30 and tp=70, then the equation reduces to (approximately - the "0.13" is actually 0.133333333333.... , and the "0.87" 1 minus that!):

Ct = 0.87 + 0.13 * ( (Ib/It)² )

So, if Ib=It (such that it is assumed that conductor is running at ‘maximum operating temperature’), then Ct = 1 (i.e. use the tabulated 70° VD figures).

At the other extreme, as Ib approaches zero (approaches no current flowing, so conductor at ambient temperature), then Ct approaches 0.87, so use 0.87 times the tabulated VD figures when current is so low that cable is at ambient temp (30°).

Kind Regards, John

 

[JohnW2]

1mm² is 44, 10mm² is 4.4 so can we assume they are correct? 44 / 2.5 = 17.6 ... Perhaps the VD figures are just not accurate enough

I suppose, strictly speaking, you should say 'not precise enough'. Most of the VD figures (and all of the VD figures for CSAs up to 16mm²) in the BGB are given to 2 significant figures, hence 17.6 becomes 18 in the BGB. Furthermore, the 44 and 4.4 probably aren't exactly 44 and 4.4 (before reducing to 2 signif. figures), either - so your 17.6 is probably not exactly right, either.

Kind Regards, John

 

[EFLImpudence]

I've also calculated if from values in another source and it comes to 17.376mV/A/m

 

[EFLImpudence]

What is the second value that appears under the original mV/A/m?

 

[JohnW2]

I've also calculated if from values in another source and it comes to 17.376mV/A/m

At what temperature?

The IET Electrical Installation Design Guide gives a figure of 14.82 mv/A/m at 20° for 2 x 2.5mm² copper conductors, and a temp coefficient of resistance of 0.00393 per degree. By my reckoning, that comes to 17.73 mv/A/m at 70°.

All this interest in high precision is, of course, totally academic given the enormous tolerances and 'safety margins' seem to be built into most ofthe figures we work with!

Kind Regards, John

 

[JohnW2]

What is the second value that appears under the original mV/A/m?

Where? In eric's calculator? (if so, it's probably the 'original mV/A/m times Ct). However, as per above posts, it looks as if he now has to mend the equation he's using to calculate Ct - so it presumably won't currently give the right answer.

Kind Regards, John

 

[EFLImpudence]

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

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.