Maximum Cable Length??

[videz6]

can someone please help me with this question

There is a 40A BS88 fuse, protecting a 10mm cable with a 4mm Earth, it is wired to Reference method C, supply is TNS and it is protected by a 30mA RCD. What is the maximum length the cable can be?

(thats all the info i have! I haven't been to collage for couple of years now and cant remember how to do it )

 

[Sav]

Hi,
Off the top of my head I would say approx 45 meters .
Is it swa ??

Thanks and regards,
Sav

 

[ezzzekiel]

find max zs then deduct ze wich will leave r1+r2 rearange formula to make length the subject:

r=mva x length/1000

so

l=r x 1000/ mva

that should work

 

[videz6]

thanks for that mate, but i dont know if it swa. how would i work it out? is it by doing the volt drop per meter?
Thanks,
Vidz

 

[videz6]

is it

45(amps) x 4.4 x 45(meters) / 1000 = 7.92 (which is just under the maximum 4% voltdrop)

is this correct?

sorry guys but i'm a bit rusty lol

Thanks

 

[mogga]

10mm with 4mm earth says TWE to me

 

[ezzzekiel]

Dont have regs to hand but guessing something like :

Ze = 0.8
Zs max for 40a bs88 = 1.35 deduct ze zs leaves r1+r2
1.35 - 0.8 - r1+r2 of 0.55
so r1+r2 = mva (for 10mm.4mm) x length (not known) / 1000
0.55 = 6.44 x L / 1000
rearrange to length = r1+r2 x 1000 / mva for 10mm.4mm

L = R X 1000/ MVA
L = 0.55 X 1000/ 6.44 = 85 METERS

thats what the max length for zs comes out at although a volt drop calc would need to be done also

 

[videz6]

thanks for that ezzzekiel, but were did u find 6.44 for the mva, im finding 4.4 in table 4D1B

 

[ezzzekiel]

values of resistence table 9a osg.



that would give the max cable length to meet earth loop value but would be too long to meet allowed vd

 

[Sintra]

Maximum length according to volt drop on t&e cable carrying the full nominal current rating of the Overcurrent Protective Device at a nominal voltage of 230v assuming cables are not grouped.

1.0mm / 6A : 26m
1.5mm / 6A : 39m
2.5mm / 16A : 39m
2.5mm / 20A : 31m
4.0mm / 20A : 52m
4.0mm / 25A : 41m
6.0mm / 32A : 49m
10.0mm / 40A : 65m
16.0mm / 60A : 68m

 

[videz6]

im looking at table 6D2 page 130 OSG, but ur right. too long for volt drop. u sure thats the correct formula?

 

[ezzzekiel]

just looked it up and its same formula from a simmilar worked example used in "practical guide to testing book"

zmax for fuse - the ze = R1+ R2

R1+R2 x 1000/ mva = length (Rearranged from R= mva x length/ 1000)



Cant see any other way of calculating without knowing the vd or the length.
Sintra gives a value of 10.0mm / 40A : 65m but need to see how he calculated it

 

[videz6]

cool, thanks for that mate.
vidz

 

[Sintra]

(mV/A/M) * Ib * L / 1000 = VD

Rearanged.

VD / Ib / (mV/A/M) * 1000 = L

Max Vd = 11.5 V

Ib = 40A

mV/A/M for 10mm T&E = 4.4

11.5/40/4.4*1000 = 65.34M rounded down to 65M

 

[videz6]

thanks sintra, but i thought the max volt drop was 9.2 (4% of 230v)

 

[Sintra]

VOLTAGE DROP

The maximum permitted volt drop measured from the incoming supply is 3% for lighting circuits and 5% for all other circuits. p358

230v x 3% = 6.9v

230v x 5% = 11.5v

 

[JUD]

Thought I was going mad for a minute and somebody had removed my post.

This question was also posted HERE which is where I replied to it.

Ring circuit

4 x (Max Zs - Ze) ÷ ((R"1 + R"2) x Cr)


Radial

(Max Zs - Ze) ÷ ((R"1 + R"2) x Cr)


R"1 + R"2 = Resistance per meter of Line & Cpc conductors
Cr = Correction factor for conductor operating temp of 70°C

So lets assume your Ze is 0.8 (max for TN-S). The max Zs for a 40A BS88 fuse is 1.35Ω and the R"1 + R"2 of 10/4.0mm² cable is 6.44mΩ per meter.

The calculation would be:

(1.35 - 0.8) ÷ ((6.44/1000) * 1.2) = 71m

However most maximum circuit lengths are limited by Voltage Drop and the voltage drop of 71m of 10mm² cable carrying 40A is 12.5V which exceeds the 11.5V (5%) maximum VD for a power circuit.

VD = mV/A/m x Length x Current
VD = (4.4/1000) x 71 x 40 = 12.5V


So you would need to work out the max length for Voltage Drop

Ring

4 x (Max VD x 1000) ÷ (mV/A/m x Current)


Radial

(Max VD x 1000) ÷ (mV/A/m x Current)


So for your circuit the calculation for VD would be

(11.5 x 1000) ÷ (4.4 x 40) = 65m

 

[pushrod]

Hey JUD how come you're not solving my problems HERE

 

[JUD]

Let me sleep on it

 

[ezzzekiel]

ah never thought to use max allowed vd, top marks to sintra