I am making a detailed voltage drop calc in excel and I have a query in relation to the formula in Appendix 4, Section 6.1 of BS7671 for calculating conductor operating temperature correction factor Ct for volt drop calcs :

Ct = 230 + Tp - (Ca2 x Cg2 - Ib2/It2) x (Tp - 30)) / (230 + Tp)

where :
Ct = Temperature Correction Factor to account for the actual temperature reached with load current flowing taking into account the heating effect of grouping and ambient temperature.
Tp = Maximum Operating Temperature of the conductor (for PVC = 70 deg C and for XLPE = 90 deg C).
Ca = Ambient Temperature Correction Factor.
Cg = Grouping Correction Factor.
Ib = Design Load Current (in Amps).
It = Tabulated Continuous Current Carrying Capacity or Current Rating of Conductor at 30 deg C (in Amps).
30 = The Ambient Temperature (in deg C) the tabulated cable current ratings are designed to.

For cables installed in underground ducts (installation method D), the current carrying capacities are calculated using an ambient ground temperature of 20 deg C.

So for calculating the Ct for cables in below ground ducts, should the Ct factor have (Tp - 20) instead of (Tp - 30), i.e. so the formula becomes :

Ct = (230 + Tp - (Ca2 x Cg2 - Ib2/It2) x (Tp - 20)) / (230 + Tp)

For example, say I'm installing a Copper/PVC cable in underground duct, so Table 4D4A & B are applicable: In Table 4D4A & B the current ratings and volt drop values are for an ambient ground temperature of 20 deg C (i.e. if the above ground ambient air temperature is 30 deg C, the ambient temperature in duct below ground would be lower at 20 deg C).

So I'm figuring that to apply the Ct correction factor to the mV/A/M values derived from Table 4D4B for cable in underground duct then the appearance of 30 deg C in this Ct formula would be meangingless as we don't have 30 deg there to start with, we have 20 deg C and the mv/A/M values are for a conductor temperature of 70 deg C in an ambient temp of 20 deg C, so if the ambient temp in duct changes to a different temperature then Tp - 20 would seem more appropriate than Tp - 30 ??


So I think the general formula should have (Tp - Tr) instead of (Tp - 30), i.e. :

Ct = 230 + Tp - ((Ca2 x Cg2 - Ib2/It2) x (Tp - Tr)) / (230 + Tp)

Tr being the temperature the current carrying capacity values are calculated at, which for above ground installations is 30 deg C and in ducts below ground is 20 deg C.

(Tp - Tr) is the temperature rise above ambient, so for calculations of voltage drop for cables in below ground ducts I would consider the temperature rise above ambient to be (Tp - 20) rather than (Tp - 30).

I have emailed IEE via IET but haven't heard back from them yet, maybe they are on summer break ??
Regards
Barry

The note to 6.1 tells you that Gilly.

If you have an ambient of 20 and tabultaed data at 20 then you would replace 30 with Tr as you suggest but only if the ambient is 20 as well.

Be aware that protection against short circuit only and protection against short circuit and overload at lower ambient temperatures will differ as you have more headroom as a starting point for short circuit faults.

regards

OMS

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Failure is always an option

I don't see that in the note OMS. But I just have the 17th edition, are you reading from the new 18th edition, I wasnt aware it was published yet ?

It seems to be in my 17th Ed - just above the "NOTE:" about the 0.004 co-efficient. "This equation applies only ... and where the actual ambient temperature is equal to or greater than 30 deg. C" - which seems to match your deduction that "the appearance of 30 deg C in this Ct formula would be meangingless as we don't have 30 deg there to start with".
- Andy.

Yes I have that note:
"This equation applies only where the overcurrent protective device is other than a BS3036 fuse and where the actual ambient temperature is equal to or greater than 30 deg C".
but I can't see how this could be telling us to replace 30 with 20 ??

Mind you, after reading that note another query springs to mind (I better not read any more I hear ye say :-), I can't understand why the formula can't be used for lower ambient temperature than 30 deg C. If say the ambient air temperature was 15 deg C the Ca part of the formula would take care of this and give a lower Ct factor which ultimately will give a lower conductor operating temperature ?

There is no 18th edition Gilly - amendment 1 to BS 7671:2008 (2011)is the current version - AKA the big green bumper book of fun -

Just prior to the note the text tells you the equation is based on actual ambient temperature equal to or greater than 30C

the note says "for convenience, the above equation is based on the approximate temperature resistance coefficient of 0.004 per degree C at 20c for copper and aluminium conductors"

If you are creating a spradsheet to do this remember it's only the rsistive component that's temperature dependant - reactance won't be affected by temperature

Regards

OMS

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Failure is always an option

Yes OMS I am aware that Ct only applies to Rc and not Xc.
BTW what does OMS stand for, just curious.

it killed the cat, old son -

regards

OMS

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Failure is always an option

Dead right

But nice to reply to a name rather than OMS......

Can't spell Ohms????????????????
sorry couldn't resist.

OMS does put up rather a lot of resistance to having his true identity made public

It matters not - judge him by what he says, rather by who you think it might be, and you won't go far wrong.

- Andy.

What r u hiding OMS....WHO ARE YOU ....perhaps the PM

product of our fractured education system Marc -

Regards

OMS

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Failure is always an option

It's not what I'm hiding - it's what I'm hiding from maybe, Gilly -

It's not important - up to you to decide if what I post on here is in some way useful to you - or complete *****

Good luck with the spreadsheet - when you have it working for temperature you can then add load power factor.

just keep in mind that simples is best, if it needs more complexity that costs effort and time and you may well spend more in effort than you gain in a reduced conductor size - particularly if you factor in whole life costs and I2R analysis of losses.

Perhaps something for your next tab on the spreadsheet - tabulate conductor size against cable cost, and installation cost, set up for I2r loss and energy cost and set up a graph to show the crossing points - so when someone says to you why do have a 240mm2 in the job when a 185mm2 would do it, you can look them in the eye and tell them that it's cheaper to own in the long term and will save 1.73 Polar bears over 25 years.

regards

OMS

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Failure is always an option

Hi OMS
In my Voltage Drop calc spreadsheet I have included correction for power factor, conductor operating temperature.
I have also included an iteration formula for reduced load current due to the impedance of the cable in question, for example if we have a 3 phase heater rated at 20kw @ 400V the design or rated load current is 20000/(1.732 x 230) = 28.87A, this is the current that would flow if 400V was present at the heater terminals, but with a 50m run of 4mm2 copper PVC/PVC cable there is an extra 0.266 ohms in series with heater which brings down the current to 27.94A, which then has a slight less value for voltage drop.....It all helps in obtaining a more accurate calculation....I hate approximations !!

Would it be simpler to calculate the resistance of the load, add that to the impedance of the supply (including circuit conductors) and directly calculate the current from that? (Iteration brings its own problems and necessarily stops when 'close enough' rather than exactly right)
- Andy.

Really ? - how are you determining circuit length then - and how confident are you the installer will install mm perfect to the route you pick.

Ambient temperature - now that's a bit of a sod to be accurate over, diurnal swings, temperature gradients within spaces ?

May I suggest to you that for most aspects of circuit design, if you are into more than one decimal place, then you are wasting your effort - be rigorous yes, not regimentally insane

regards

OMS

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Failure is always an option

Not wanting to interfere with the current discussion but thought I would show you a very accurate way to determine hope it helps a little, if you try doing a comparison between this simple method and your method you will find very little difference

Facts to remember are the resistivity of copper is a constant based on 25 deg c for this method and that for each 1 deg C rise above 25 deg C the VD will rise 0.4%
only for cables up to 16mm sq

Example 70 meters of 2.5mm SQ MM having a supply voltage of 48 volts and a load current of say 1.35 amps at 50 deg temp.

(70 meters x1.35 amps x0.0178 resistivity)/2.5mm sq = 6.7284 @ 25 deg c
@ 50 deg = (6.7284x0.4%) x the additional 25 deg above the calculated @ 25 deg plus the original VD = 7.40124

We tested this method at the given temp and at load as indicated above and it was very accurate

Regardless of all the factors that need to be considered with the IEE regulations if you calculate the actual true temperature your cable will subjected to the simple formula works and has proved here to be very accurate

Hope it helps a little in your quest