Maximum Cable Length

[JUD]

Ok. I've been trying to get my head round this all day but I'm getting nowhere.

On page 45 of the OSG (Table 7.1) it says the maximum length of a lighting circuit wired in 1.5/1.0 cable on a 6 amp Type B MCB is 90 meters.

However, working out the voltage drop for this length of cable I'm getting 13.76 Volts (after correction for operating temp) which is double the allowable VD for a lighting circuit.

Am I missing something?? Does anybody know how the values in the OSG are worked out??

Any help GREATLY appreciated!!

JUD

 

[roukel01]

the onsite guide has a specific 'Lighting' table which is what you are looking at, if you look it says "load evenly distributed" so you can't simply caluculate the voltage drop. All of the run isn't carrying the full load and so voltage drop along the circuit will differ.
Only the initial run will be carrying the full load, from there on, the load reduces as each light is fed.

 

[JUD]

Thanks roukel01.

After doing a bit of digging last night I came across this which supports what you have said.

the bit you are forgetting is the little comment in bold just at the top of the table...

Lighting circuits (3% voltage drop load distributed)

"Load Distributed" is the key point.

IF the whole of your 6 amps was right at the end of the 90m cable then you are correct volt drop would be 15+volts.

BUT... lets assume the load is distubuted... try a few easy numbers

Consider if the lighting circuit has 12 x 100watt light bulbs (stuff the green energy saving lark for the moment!)

AND lets say they are equally spaced along the cable run in pairs,
(could be 6 rooms each with 2 bulbs in)

so 90m length divided by 6 is 15m sections.

15m then 200w load [call this bit A]
30m then 200w load [call this bit B]
45m then 200w load [call this bit C]
60m then 200w load [call this bit D]
75m then 200w load [call this bit E]
90m then 200w load [call this bit F]


so bit F is going to carry 200w 0.87A & thus drop 0.38v (29x0.87x15)/1000.
so bit E is going to carry 400w 1.74 & thus drop 0.76v (29x1.74x15)/1000.
so bit D is going to carry 600w 2.61A & thus drop 1.13v (29x2.61x15)/1000.
so bit C is going to carry 800w 3.48A & thus drop 1.51v (29x3.48x15)/1000.
so bit B is going to carry 1000w 4.35A & thus drop 1.89v (29x4.35x15)/1000.
and
bit A is going to carry 1200w 5.22A & thus drop 2.27v (29x5.22x15)/1000.

Add all these volt drops together and you get somewhere around 7.94v

which is still bigger than the 6.9v max that you correctly mention.

But we still haven't taken account of diversity.
e.g. we have calculated EVERY light is on all the time.

if we assume the 66% diversity rule of thumb with lighting circuits
then it now drops down to approx 5.25v
Jobs a goodun as they say!!

Hope this all makes sense..

I don't know if this is how the IEE do their calcs? but they are bits that need to be taken into account.

Also don't forget... real world probably hasn't got all 100w bulbs!
chuck a few 60w / 40w, or low energy lamps in and you get numbers down to 3v 1.7%

JUD