# how to find out the maximum permissible length for a spur .

how to find out the maximum permissible length for a spur using mathematics.

this is how i learnt how to calculate volt drop

TYBE B CIRCUIT BREAKER 32 AMP

LIVE 2.5 MM

CPC 1.5 MM

length 66 mtr

R1 + R2

0.32 OHMS

RN

rn for 2.5 mm is 7.41

7.41 * 66 mtrs / 1000 = .49

RN = 0.49

MAx Zs for type B 32 amp circuit breaker is 1.44

230 v / 160 amps = 1.4375

if you want to add a spur to this circuit, how much wire is permissible?
mrsonic
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collectors
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As collectors says length of any circuit not only limited by V-D but also Zs and temperature ect.
Not sure I follow your reasoning as you talk of Volt drop but then mention Zs.
One implies measuring R1+R2 the other R1+Rn.
Your question at the end of how long can the spur be must depend upon:
a] Zs at the point you spur from subtracted from allowable Zs in tables

eg say Zs 0.6 measured gives 1.44-0.4= 0.84 which if divided by resistance per metre of 0.01951 ohms (from table 1a of GN3 app.1)
= 43Metres.
b] max allowed volt drop which is more difficult and relies upon some assumptions eg existing load at point of spur Say 20A plus potential load on spur say 8A?

To calculate V-D you could disconnect L+N at board and connect 4 ends together, measure resistance at spur off point, use ohms law to calculate V-D at that point, for 28A load
then using 0.0148ohms from GN1 again as 2x2.5 cores / metre calc based on 8A load the additional length allowed by the remaining V-D .

not sure I've said that understandably (but I know what I mean!!!)
OR you could just keep the ring inside 100Msq including the spur as per APPENDIX 15 of BRBook of fun!
regards Sparx (hung over)!
sparx
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You say spur and 433.2.2 is the regulation that controls the max length of a spur and it has nothing to do with calculations.

The device protecting a conductor against overload may be installed along the run of that conductor if the part of the run between the point where a change occurs (in cross-sectional area, method of installation, type of cable or conductor. or in environmental conditions) and the position of the protective device has neither branch circuits nor outlets for connection of current-using equipment and fulfils at least one of the following conditions:
(i) It is protected against fault current in accordance with the requirements stated in Section 434
(ii) Its length does not exceed 3 m, it is installed in such a manner as to reduce the risk of fault to a minimum, and it is installed in such a manner as to reduce to a minimum the risk of fire or danger to persons (see also Regulation 434.2.1).

So since the "device protecting a conductor against overload" is the 13A fuse in the plug or FCU the length of a spur is limited to 3 meters.

If you use a 13A FCU at point where you connect to ring then you have formed a new circuit and it is a radial not a spur. However then you are looking at 2.42 ohms not 1.44 ohms.

However we do not normally measure the prospective short circuit current in ohms but normally measured in amps OK I know they are related but it is after all the amps that matter.

The line / earth impedance needs to be better than 200 ohms to show it is reliable enough but in real terms the line / earth should be considered in relationship of earthing system and a figure that shows it is likely to be sound if what is looked for which will be within 2 ohms of the Ze.

With RCD protection on all sockets under 20A we are now only interested in the line / neutral figures both so the magnetic part of the trip will operate and that the volt drop is within the 5% allowed.

Page 258 of BS7671:2008 does give the formula however one would hope one would not be trying to cut it so fine and although we can use just over 100 meters of 2.5mm in a standard 32A ring main to cut it that fine is asking for problems where the PSC current is measured.

5% = 9.2 volts and with 13A on 2.5mm cable with the corrected mV/A/m of 16.5 then max length for volt drop is 42.5 meters. However I would not want to cut it so fine and would say 35 meters max.

As to 13A not sure that is figure used as we are allowed a double socket so may find the ratting is higher for a spur but since a spur is limited to 3 meters anyway why bother calculating? Well I have it all set up in Excel so did it just for interest and 25.8 meters if you consider 20A as design current.

I made the Excel program a long time ago when the new BS7671:2008 came out and we were debating max cable length in a ring (106 meters) and it was pointed out now dependent on volt drop. However I can't remember now exactly what I allowed for with calculations.
ericmark
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