Discuss 300M Supply to barns in the Commercial Electrical Advice area at ElectriciansForums.net

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fjr1300

Hi All

I’ve got to install electricity in 7 adjoining barns (10M*17M each)
Each barn will have 2*(Double waterproof Fluorescent lights) and 5 (58W) heated water troughs making a load of 438W per barn.

The loads are single phase 230V and will come from a 3 phase supply – load balancing 2 barns per phase.
The supply is around 250M away and will be underground so Method D – 4 core cable.

Now I'm not sure of which column to use on tables 4D4A and 4D4B :confused:
(Pages 280-281 17tg regs) because of the single loads but using 4 core 3 phase cables.

I'm doing the installation in France where they use non-armoured cable (less weight thank god :)) (60-80cm deep in ground) but assume the armoured tables would be ok.

I obviously would not use 1.5mm cable :D because the volt drop is too high and I think I'd allow for 16Amps load per phase (3680W) so that power sockets could be fitted but I'd like to see if I've got the correction factors and tables right.

I don't want to oversize the cable since the weight of 250M-300M, 4 core, is going to be fun installing :eek:.

Thanks

Leigh


Calculations:-

Voltage = 230V

Max Voltage drop in is 4% = 9.2V

Length of Cable 300 (250M + approx middle of barns)


Required Power per barn 434 watts

Since there are 7 Barns, some of then will share the same phase
Max power, IE 2 Barns 868 Watts



Design Current 3.77 Ib Amps


Selected Breaker 5 In Amps
(Manual Selection)


Factors
1 Ca Equals 1 in 30Deg C Temp
0.65 Cg 4 Circuits (L1/L2/L3/nN)
1 Ci Equals 1 since in Ground and not Insulation involved
0.9 Cc Equals 0.9 since in ground


Cable Current Capacity 8.55 Amps It
It ≥ In / (Ca * Cg * Ci * Cc)


Select Cable size that has current capactiy ≥ to It
1.5mm
(From Table 4D4A Page 280 17th Regs)

From Table 4D4B (4) the mV/A/m) = 29 (Single phase Col 3)


Single Phase Loads
Therefore
Voltage Drop = (mV/A/m)*Ib*L / 1000 where IB=3.77 and L = 300
32.83 Vdrop
 
Hey.

300M is one hell of a run for a sub-main.

I think I would be looking into having a new metered supply being installed by the DNO.

If this isn't possible then your calcs need to be spot on also I would seriously be cosidering future usage of the barns - Is there a possibility that they may want to covert one or two of them into living spaces in time.

If this is a possibility then the sub would need to be huge.

Even with the loading you've got at present and allowing for all factors I would be looking at 16mm as a minimum.

But I would seriously look at the new supply first.

Hope this helps.
 
Some nice person burnt a large barn down so the farmer is going to put up some temporary plastic/aluminium ones (2 years max) so they are never going to used permantly.

I'm worried that the cable will be too heavy for me to handle - even with the help of tractors ..etc
Looks like getting EDF to see what they would charge since they've got the equipment.

Not found anything in the French regs to help yet, appear to be very much like the 16th/17th.

Leigh
 
Hi Leigh.

Just thinking, these temporary structures do they have to be so far away from the supply incomer?? Could you not explain to the customer all the issues involved with having them so far away??

Only a thought.
 
i find there are some good web sites that will calculate cable size for u but looks like would need at least 16mm if its underground
 
The farmer likes to do things the hard way, he's says the nice fields next to the house are too soft for the cattle (Limosin cows - big buggers), so there's an old wood 300M away thats got to be flattened and the stumps removed.
Then there's the muddy lane which has also to be flattened because it's too steep for the cattle to walk up in parts, and has large rocks in the ground as well.

He asked if i would dig the trench - 300M, 800mm deep in rocky ground - I think not :rolleyes:

Like I say it's not a simple job :eek:

Using the above calculation methods, I'd need to used 25mm cable, which would give me a volt drop of 8.4 volts when taking 16Amps (single phase).
A normal load of 868W (2 Barns at 484W per phase) would give a 1.98V drop.

This is assuming a mV/A/m of 1.75 resistive load - which of course the fluorescent lights are not :confused:

A 10mm cable would give 3.17 Vdrop under normal loading and 13.44 Vdrop if 16Amps was taken

I'm getting to like the "get EDF to do the job" bit and I'll finish off wiring the barns internally

Leigh
 

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