?


Postby mrsonic » Tue Apr 06, 2010 9:44 pm

C TYPE 16 amp mcb

wire live 4 mm2

cpc 1.5 mm2

length of wire from Cu to shed is 35 mtrs

r1 + r2 = 16.71 per mtr x 35 mtrs / 1000 = 0.58485

rn = 4.61 x 35 mtr / 1000 =

0.16135 x 2 x 16 A =5.1632 voltage drop


would this circuit be legal if it were created?
Last edited by mrsonic on Wed Apr 07, 2010 11:17 am, edited 1 time in total.
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Postby ericmark » Tue Apr 06, 2010 10:39 pm

Legal and complying with regulations are two different things. Likely it would be legal but as complying with regulations 3% is max volt drop for lighting and unlikely the shed will have no lights.
If domestic building there is also Part P. Boats, caravans and Scotsmen don't have Part P. They are allowed to electrocute themselves!!!!!
The volt drop regulations are not too clear as to if calculated on supply available or power used. So if a shed was supplied with a 16A supply and has no sockets only a single light then although it has 16A supply one would calculate volt drop on max bulb size. However when you also have sockets then it would normally be the supply size.
Earthing is harder as it would depend on location. In middle of a housing estate and surrounded by houses likely you can use house earth but on a farm with no building beyond the shed likely it would need a TT supply.
The use of RCD protection does remove some of the dangers but even then one has to consider if with short circuit there is enough voltage to work the RCD. You may need to use an active RCD.
Sorry to be so vague but this is the problem. When one looks at a job one considers all one can see but to try to consider correct method without seeing a job would need a crystal ball. Or at least some photos and this forum does not allow them.
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Postby ericmark » Wed Apr 07, 2010 8:36 am

I though I had answered this but here goes.
C type = x10 so working on 160A before MCB will trip in time allowed.
Max voltage to earth allowed before it trips = 50
Voltage needed to make it trip 89.04
So will not pass unless RCD is used.
Max volt drop for sockets = 5% = 11.5v
Max volt drop for lights = 3% = 6.9v
Your volt drop = 1.568v so OK
With B type MCB it just passes at 44.52v but I question use of 1.5mm for earth! How? Never seen a 16mm cable with 1.5mm earth and if not part of cable then min size is 4mm.
Something wrong!
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Postby mrsonic » Wed Apr 07, 2010 7:10 pm

i heard that if the ACTUAL Zs Value is LOWER than the MAX Zs value then the circuit is okay. so 10 x 16 amps = 160 . 230 / 160 = 1.4375 max zs value. is the actual zs value added to r1 + r2 value?
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Postby mrsonic » Wed Apr 07, 2010 7:19 pm

[quote] So if a shed was supplied with a 16A supply and has no sockets only a single light then although it has 16A supply one would calculate volt drop on max bulb size[/quote]

example

60 watt bulb and 1 socket to power up 1800 watt heater.

1860 watts
volt drop for 4 mm2 = .011 per mtr
8.09 amps x 35 mtr x .011
volt drop = 3.11
Last edited by mrsonic on Sat Apr 10, 2010 4:32 pm, edited 1 time in total.
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Postby ericmark » Thu Apr 08, 2010 12:38 pm

What I was trying to say
For earth loop impedance then supply limit i.e. D16 MCB is what you go by.
For volt drop it is actual use.
However since electricians often don't have control of use then often we still go by MCB size.
However if it were for example supplying a HF florescent unit in a shed, and only that, and the unit gives a input voltage of 100 - 240v AC/DC, then to say the volt drop must be limited to 3% is daft.
And yes the ELI must be lower than the amount which would allow it to trip.
So calculation C type = x10 so max current = 160A @ 230v that's 1.4375 ohms max so assuming 0.35 (PME) on incoming supply you have 1.0875 ohms so around 6.17 meters is max length of a 4mm supply.
But you want to use a 1.5mm earth and that will give max of around 4.26 meters unless a RCD is used.
Use a B16 and your up to 9.894 meters
Use a B10 and your up to 16.65 meters
Use a B10 and 4mm for earth and up to 24 meters
Use a B10 and 6mm for earth and up to 36.38 meters
Now on a 35 meter run you are unlikely to use twin and earth much more likely to be SWA and if you use 3 core then both the SWA and the core can be used for earth and now with a B10 and 4mm you are likely to be able to get the readings required however since the SWA cross sectional area varies between makes I can't calculate the value.
Sorry to say it is not that easy in real terms as one does get parallel earths and it is the meter reading that matters.
Using a TT system you may have better results this will need RCD protection and as I said it depends on location you may have to use a TT system anyway.

Electrical installations outdoors: a supply to a detached outbuilding in Wiring Matters (Google "theiet.org/publishing/wiring-regulations/mag/2005/index.cfm") does go through all the ins and outs but do remember this was pre 2008 so there will be some changes to comply with BS7671:2008
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Postby mrsonic » Sat Apr 10, 2010 4:55 pm

thanks for the clarification.
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Postby mrsonic » Sat Apr 10, 2010 9:17 pm

"So calculation C type = x10 so max current = 160A @ 230v that's 1.4375 ohms max so assuming 0.35 (PME) on incoming supply you have 1.0875 ohms so around [b]6.17 meters is max length of a 4mm supply[/b]. "


please can you show the calculation method you used to determine the max length for 4 mm supply.

R * 1000 / mv = length?
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Postby ericmark » Sun Apr 11, 2010 1:55 pm

I normally use excel and can't remember if I used pre-saved sheet or new one when working it out. Seems I may have entered figures in wrong holes. Since all the figures given are for return volt drop I had to half the figures for 1.5mm and 4mm then add together the results.
Your 1.0875 ohms I agree with so at approx 0.02 ohms per meter which means more like 54 meters which seems to tie in with the 102 meters allowed with ring main I blame the rum!
I do remember around a year ago querying the new length for a ring main now 5% rather than 4% volt drop. And although the ring main is rated 32A only 20A is considered as being draw at the centre and the remaining 12A to be even distributed so half of that is 6A so 26A is being drawn on cables with a design current of 2 x 21A = 42A so using temperature correction instead of 0.018mv/A/m it worked out around the 0.016 which explained the longer length to what I had calculated before.
While doing this I had made a number of excel programs so I could re-calculate using 4mm and 6mm for radial and ring main. I think I had re-loaded one of these and somewhere I had the wrong figures entered.
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