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

 

[steptoe]

kk, the OSG gives you the maximum length,

if after your calcs for VD give you a different length, then you simply use the smallest length.

 

[SPECIAL LOCATION]

See this thread here...

http://www.talk.electricianforum.co.uk/showthread.php?t=7618&highlight=VOLT+DROP+LIGHT

and post #10's a good un IMHO:Blushing

it goes somthing like this....

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%

 

[steptoe]

See this thread here...http://www.talk.electricianforum.co.uk/showthread.php?t=7618&highlight=VOLT+DROP+LIGHT

and post #10's a good un IMHOBlushing

it goes somthing like this....

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%

ignore me,

this is much better

 

[JUD]

Thanks guys!!!

When you put it like that it makes perfect sense.

 

[SPECIAL LOCATION]

ignore me,this is much better

:O:O:O

How could we EVER ignore you Steps!Applaud Smiley:x

ROTFWLROTFWLGuinnessGuinnessGuinness

 

[Apache]

:O:O:OHow could we EVER ignore you Steps!Applaud Smiley:x

ROTFWLROTFWLGuinnessGuinnessGuinness

On this forum 'ignore' seems to have reversed its meaning to 'pay particular attention to'

 

[steptoe]

of course at the expense of poor Brian,

whom I think Deke has taken exception to you hijacking for your sig....

 

[Evans Electric]

See this thread here...http://www.talk.electricianforum.co.uk/showthread.php?t=7618&highlight=VOLT+DROP+LIGHT

and post #10's a good un IMHO:Blushing

it goes somthing like this....

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%

I was just about to say exactly the same but Specs types faster than me , even with those weird paws of his !!

Anyone seen Brian , he,s left me .

 

[SPECIAL LOCATION]

I was just about to say exactly the same but Specs types faster than me , even with those weird paws of his !!Anyone seen Brian , he,s left me .

I copied it off that sheet of A4 you left on the passenger seat of you van...

That'll teach you to leave the windows open!!! ] ] ] ]

Last time I saw Brian he was over at KME's giving him a back massage!

:|

 

[Evans Electric]

Is he now !! Brian as no loyalty , he'll go off with anyone, seeing as I spent a mega long "ignoring" thread trying to conjure him up , he is so fickle. He was dancing with Apache's cows the other day .

 

[Apache]

Is he now !! Brian as no loyalty , he'll go off with anyone, seeing as I spent a mega long "ignoring" thread trying to conjure him up , he is so fickle. He was dancing with Apache's cows the other day .

still is.

 

[Evans Electric]

Yes there he is , he doesn't care about me , treats my place like a hotel , he'll go off with anyone , another sparks with a shiny new van and he,s off .

He,s looking for the bright city lights and the wild side of life but he'll be back when the bright lights grow dim.

Yes you'll walk the floor ... the way I do , :coat

Your Cheatin' Heart is gonna tell on you . ]

Who sang that , EXBBC ??

 

[ADS]

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.

This is good stuff, thank god we don't have to do it for every job

The only thing that doesn't add up (and I'm being picky now) is the values of current that you have used.

You've used 230v each time that you have calculated the current drawn at each set of lights - this won't be the case as, as you correctly stated, the voltage will drop by a certain amount every 15 metres.

For example - Resistance of the bulbs remains constant (264ohms per pair) so, if the volt drop by the end of bit E was indeed 7.56v (total), then the current drawn through bit F would be 0.84 A, not 0.87.

This would also apply to each other bit as you moved along the circuit.

Of course, this lower current drawn will in turn affect the volt drop, so a nasty little circle - my head is spinning? :|

I think I'd turn all the lights on and measure it ha ha

Does what I've said make sense?

 

[SPECIAL LOCATION]

This is good stuff, thank god we don't have to do it for every jobThe only thing that doesn't add up (and I'm being picky now) is the values of current that you have used.

You've used 230v each time that you have calculated the current drawn at each set of lights - this won't be the case as, as you correctly stated, the voltage will drop by a certain amount every 15 metres.

For example - Resistance of the bulbs remains constant (264ohms per pair) so, if the volt drop by the end of bit E was indeed 7.56v (total), then the current drawn through bit F would be 0.84 A, not 0.87.

This would also apply to each other bit as you moved along the circuit.

Of course, this lower current drawn will in turn affect the volt drop, so a nasty little circle - my head is spinning ?:|

I think I'd turn all the lights on and measure it ha ha

Does what I've said make sense?

Bottom line..... :_|

Yes you are being Picky!!!:^OROTFWLROTFWLROTFWL

But quite rightly it again helps illustrate the concept of volt drop with a distributed load is a little more complex than just a simple x+y=z calculation!

Whereas a Cooker, Shower, Immersion heater etc.. radial circuit is just a Load at the end of the circuit.

Sockets or lighting radials are a slightly different beast. :|

Now if we really want to get into 'head spinning'....

How about the actual real voltdrop on a ring circuit!!!

Anoraks on

Zips up

Hood strings pulled tight!

Start with a 100m 2.5mm/1.5mmCPC T&E ring circuit....

Try a single 1kW load exactly half way round the ring (50m point)

Then hows about if that 1kW load is Quarter or a Third of the way around? (25m or 33.3m)

Now lets add in more loads what about 3 x 1kW loads at 25m, 33.3m & 50m form one end off the ring?

Or try working:-

1500w load @ 25m

1000w load @ 33.3m

2500w load @ 50m

400w load @ 66.6m

2000w load @ 75m

total load 7400w approx 32A

But what IS the volt-drop.. That would get your head spinning!

bad day explode

And which way round the ring are all these currents going?

; \

Anyone got a really big Anorak? :Blushing

Could possibly be a Duffle Coat job?????

]

:coat

 

[JUD]

Also the voltage drop values from 7671 are calculated on the cables maximum operating temperature so if you correct the values using the formula on page 258 of the BRB the voltage drops become even lower.

Works out something like this:

See this thread here...http://www.talk.electricianforum.co.uk/showthread.php?t=7618&highlight=VOLT+DROP+LIGHT

and post #10's a good un IMHO:Blushing

it goes somthing like this....

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 0.89A & thus drop 0.34v (25.1x0.89x15)/1000.

so bit E is going to carry 400w 1.78 & thus drop 0.67v (25.1x1.78x15)/1000.

so bit D is going to carry 600w 2.67A & thus drop 1.00v (25.1x2.67x15)/1000.

so bit C is going to carry 800w 3.53A & thus drop 1.34v (25.3x3.53x15)/1000.

so bit B is going to carry 1000w 4.39A & thus drop 1.67v (25.3x4.39x15)/1000.

so bit A is going to carry 1200w 5.22A & thus drop 1.99v (25.4x5.22x15)/1000.

Add all these volt drops together and you get somewhere around 7.01v

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 4.62v

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%

Since we're being picky

 

[ADS]

Or try working:-

1500w load @ 25m

1000w load @ 33.3m

2500w load @ 50m

400w load @ 66.6m

2000w load @ 75m

total load 7400w approx 32A

But what IS the volt-drop.. That would get your head spinning!

bad day explode

And which way round the ring are all these currents going?

; \

Not a chance, mate, you did my head in with the last one - I'm not even gonna start thinking about this one:^O

 

[SPECIAL LOCATION]

Not a chance, mate, you did my head in with the last one - I'm not even gonna start thinking about this one:^O

You little B* you!!!! X(

I look and see new post ADS..

me thinks oh great ADS has got some really great easy to follow calcs for sorting ring volt drop...................

But NO! :_| :_| :_| :_| :_|

hopes dashed again!

you know how to break a mans heart you do.. :|

I may even complain to Admin! :Blushing

:^O :^O:^O

ROTFWLROTFWLROTFWLROTFWLROTFWLROTFWL

:coffee