Cable To Shed

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Thanks ProDave - this is how we were working out the overall breakdown (which is probably excessive but more worse case scenario):-

Exterior Lighting  - max 2kW total:-

 - 2 x 200w flood lights

 - 2 x 100w up/down lights

 - 16 x 50w uplighters (run on 12v transformers)

In Outbuilding - max 10kW total:-

- Heater 2000w

- Kettle 3000w

- PC 150w

- Power tools x 2  3200w (1600w each)

- Lights inside shed - 150w

- other small electrical items - 200w

Think you're right about running a dedicated network cable down as we'll have a trench anyway!

Special Location: the lights furthest from the shed are only the 12v 50w, the flood lights will be on/near the shed.
You are forgetting to apply diversity.

Diversity basically allows you to down rate a circuit on the basis that not everything is on at the same time.

For instance it would be very unlikely to have the kettle on at the same time as everything else. you STOP working for a tea break!!

So I would think a 3KW radial socket circuit would suffice, and look long and hard at your lighting requirements.  I know they are "old fashioned" but one or two 58W 5ft flourescents would give you plenty of light in a shed, and low energy outside lighting as well.

 
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on saying that Dave,

in my shed are 4x58w fittings, and yes, sometimes all on at once,!

along with plug in lamps as well sometimes.

oh, only a 'small' 12x8 ,

I couldnt possible put anything bigger in the space and still remain married, some may say I missed a chance there.  :|

I dont think the in-shed lighting is an issue, more the VD for the outside lights,

oh, remember to include the beer fridge in your calcs.  :D

 
"I'll just warm the garden office up (3kW) so it's nice and warm for when I log onto Redtube "work" and finish that report! Whilst it's warming up I'll finish that bit of shredding (2.5kW) under the glare of the two lights (1kW)".

It all mounts up!

Not a real life scenario or indeed anyone I know!

:Blushing

 
Can't see you using more than 30A  and at 75 mtrs I'd be looking at a 10mm X 3 core SWA or a  2 core if you TT it .

All  those 50W lamps don't help TBH .

 
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Not with you there Steve...............I'm using a design current  (Ib)  of 30A .   VD is within parameters . 

A 16mm working at its FLC of 78A  would drop 16V    but I'm designing for a 30A supply . 

 
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Well Steve ,  either Amtrack has a clitch in it or my maths do .     I know its a good length at 75m  but 25mm can't be right ........can it ?   Am I getting past it ? Don't bother to answer that .

 
Well Deke's BOFP method is showing  a 10mm  with a load of 30A  dropping 9.9 V which is less than the allowed 5% leaving 220V   ( Using the "Lets pretend the voltage is only 230v when in the real world its over 240 v  method) .

 
your limit is 3% as you have lighting out there.  

 
Supply voltage = 230 Volts
Ib - design current = 30 Amps
Protective device type = MCB type B
In - protective device rating = 32 Amps
Length of run of cable = 70 metres
Power factor = 1

Cable Type : Thermosetting 90°C 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 = 32 / (1 x 1 x 1 x 0.9 )
It = 35.56 Amps
From TABLE 4E4A Cable selected = 4 mm²

TABLE  4E4B For 4 mm² ; mV/A/m  =   12
mV/A/m corrected for power factor = mV/A/m x Power Factor = 12 x 1 = 12

Voltdrop  =  (mV/A/m   x   Length  x  Design current) / 1000
Voltdrop   =  (  12   x   70   x   30  ) / 1000
Voltdrop   =  25.2  Volts
(Maximum permissible voltdrop (regulatuon - 525) = 6.9 Volts)
This exceeds the maximum voltdrop, so we have to Increase cable size and recalculate

TABLE  4E4B For 16 mm² ; mV/A/m  =   2.9
mV/A/m corrected for power factor = mV/A/m x Power Factor = 2.9 x 1 = 2.9
Voltdrop  =  (mV/A/m   x   Length  x  Design current) / 1000
Voltdrop   =  (  2.9   x   70   x   30  ) / 1000
Voltdrop   =  6.09  Volts

Calculated Cable size = 16 mm²,       Minimum Earth conductor size = 16 mm²  (Table 54.7)

Maximum Cable Length = 79.3 Metres


Any help ? :)

 
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