Cable Size Help

[mk1rob]

Hi guys

A customer has asked what cable size he would need for an out building in his back garden. In the out building, he's planning on having 3 or 4 general use sockets, 2 lights inside and 6 pillar lights down the garden. The cable length needed to the consumer unit is 85m and will be SWA. I'm thinking 16mm 3core because of the distance but was wondering what your thoughts were? Maybe 10mm would be big enough?

Thanks

 

[Bruspark]

Get out your tables and a calculator ?

 

[ProDave]

What do you think your maximum load will be?

 

[mk1rob]

The customer has told me the sockets are just for general use and nothing specific so i would imagine no more than 16A.

 

[m4tty]

Supply voltage = 230 Volts
Ib - design current = 16 Amps
Protective device type = MCB type B
In - protective device rating = 16 Amps
Length of run of cable = 85 metres
Power factor = 1

Cable Type : General purpose PVC ARMOURED - Multicore

Installation Method : Sheathed, armoured or multicore cables direct in the ground: without added mechanical protection.
method 0

Ambient temp = 20  °C
Number of circuits including this one = 1
Length of cable in thermal insulation = none

Apply Correction factors:
From TABLE 4C2 : Cg = 1 (Grouping)
From TABLE 4B2 : Ca = 1 (Ambient temp)   -  Ground Temperature  : 20 °C
From TABLE 52.2 : Ci = 1 (Insulation)
Protective device factor  : Cc = 0.9 (Burried direct)

It = tabulated current carrying capacity
It = In / (Cg x Ci x Ca x Cc)
It = 16 / (1 x 1 x 1 x 0.9 )
It = 17.78 Amps
From TABLE 4D4A Cable selected = 1.5 mm²

TABLE  4D4B For 1.5 mm² ; mV/A/m  =   29
mV/A/m corrected for power factor = mV/A/m x Power Factor = 29 x 1 = 29

Voltdrop  =  (mV/A/m   x   Length  x  Design current) / 1000
Voltdrop   =  (  29   x   85   x   16  ) / 1000
Voltdrop   =  39.44  Volts
(Maximum permissible voltdrop (regulatuon - 525) = 6.9 Volts)
This exceeds the maximum voltdrop, so we have to Increase cable size and recalculate

TABLE  4D4B For 10 mm² ; mV/A/m  =   4.4
mV/A/m corrected for power factor = mV/A/m x Power Factor = 4.4 x 1 = 4.4
Voltdrop  =  (mV/A/m   x   Length  x  Design current) / 1000
Voltdrop   =  (  4.4   x   85   x   16  ) / 1000
Voltdrop   =  5.98  Volts

Calculated Cable size = 10 mm²,       Minimum Earth conductor size = 10 mm²  (Table 54.7)

Maximum Cable Length = 98.0 Metres


might help ??

 

[Sidewinder]

Design EFLI?...

 

[Bruspark]

Any other heating plans that would persuade you there won't be 2 x 2KW electric heaters running on top of the garden lights and the kettle . . .

 

[apprentice87]

Well said Paul!!

john...

 

[m4tty]

And if app87's round there. Add a few welders into the equation

 

[mk1rob]

Thanks for the replies everyone. M4tty thats great thanks! Looks like 10mm will do it then! However think i might go for the 16mm and overkill it. I guess that way the customer can plug what he likes in. Thanks again, very helpful!

 

[Blue Duck]

Thanks for the replies everyone. M4tty thats great thanks! Looks like 10mm will do it then! However think i might go for the 16mm and overkill it. I guess that way the customer can plug what he likes in. Thanks again, very helpful!

Think you should give m4tty 50% of the money as he's done 50% of the work for you!

Nice one m4tty!!!

 

[soulman]

Hi, Matty, thank you for that calculation is was good piece of advice.  where did obtain the figure 0.9 for buried direct?  Also where did you get maximum circuit length from? Are you using the 3% VD.as lights on the circuit.  

Also if you dont mind could you check this calculation for efli please, as sidewinder has suggested & tell me where i am going wrong.

10mm Line = 1.83mohms/m  10mm E.C 1.83mohms/m  =3.66mhoms/m plus 1.2 for increase temp = 4.39m/ohms/m

Protective device In = 16A 60898 = 2.87ohms

Can we use TN-S for earthing system 0.80ohms

2.87ohms - 0.80ohms = 2.07ohms

2.07 x 1000/4.39 = 471m

471m 

V.D would be 33v

So am i correct in saying in theory, if volt drop was o.k eli would be fine.  Or am i way off the mark?

Cheers