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Topic Title: Cable calculations to BS7671:2008
Topic Summary: Anyone like them, want to practice some?
Created On: 24 March 2008 03:18 PM
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 24 March 2008 03:18 PM
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bakey1959

Posts: 767
Joined: 24 October 2006

I would appreciate some reviews of the following cable selection procedures taken from BS7671:2008, and if anyone is so inclined, your results of three examples at the end.

If anyone wants it in word format private email your address and I will forward (better layout than this pasted version).

CURRENT-CARRYING CAPACITY AND VOLTAGE DROP FOR CABLES AND FLEXIBLE CORDS


The two basic calculations that determine the size of a cable to be used for a particular purpose are:

1. the current rating ( Iz ) of the cable under defined installation conditions

2. the maximum permitted voltage drop as defined in BS7671:2008 by Section 525 and Appendix 12.

Appendix 4 details the current carrying capacities,
correction factors and voltage drop for cables and flexible cords.




The factors which influence the cable selection include:

1. the design current ( Ib ) or rating of the equipment to be connected, and

2. the rating of the circuit protective device ( In ) , and

3. the tabulated current carrying capacity ( It ) of the cable, and

4. the installed current carrying capacity ( Iz ) of the cable, taking into account any correction factor(s) (C)



BS7671:2008 requires that: Iz ≥ In ≥ Ib






Procedure for determining the size of cable to be used.

Current carrying capacity

Divide the rating of the selected protective device ( In ) by any correction factors (C) and select a cable from Appendix 4 that is not less than this value ( It ).

It ≥ Current rating of protective device (In)
Any applicable correction factors (C)

CORRECTION FACTORS REFERENCES
Symbol Description BS7671:2008

Ca
The ambient temperature correction factor Appendix 4

Table 4B1 ( in free air )
or
Table 4B2 ( buried )

Cg
The grouping correction factor. Appendix 4

Table 4C1, 4C4 or 4C5 (in free air)
or
Table 4C2 or 4C3 ( buried or in ducts )

Cc
The protective device correction factor. Appendix 4

Section 4 gives 0.725 for
BS3036 'rewireable' fuses.


Ci
The correction factor for cables embedded in thermal insulation Regulation 523.7 - Table 52.2
Length in insulation Divide by
50mm 0.88
100mm 0.78
200mm 0.63
400mm 0.51
500mm or above 0.50


Voltage drop calculation

For 230v installations fed from a public supply the maximum permissible voltage drops are 3% (6.9v) for lighting and 5% (11.5v) for other uses.

Voltage drop = mV x Ib x L where: mV is the millivolts per amp per metre taken from tables 4D1A to 4J4A, and
Ib is design current of the circuit, and
L is the cable length in meters.

Anyone interested in working out the minimum cable sizes for the following examples?

1. A circuit has a 14kW heating load fed at 230v. The cables are thermosetting (90 degrees C) singles having copper conductors enclosed in conduit, fixed on a wall and run over a length of 36m. Overcurrent protection is provided by a BS88 fuse.
The ambient temperature is 40 degrees C and the circuit is grouped with two others of the same size, all circuits being loaded above 30%. The circuit does not run through thermal insulation.

2. A 2kW commercial lighting load is supplied at 230v. The cable length is 40m and the cable is PVC/PVC flat twin (thermoplastic 70 degree C). Overcurrent protection is provided by a Type B BS EN 60898 MCB. The ambient temperature is 30 degree C and the cable passes through and is surrounded by thermal insulation for a distance of 200mm. The cable is grouped with one other cable of the same size throughout the run and both circuits are loaded above 30%.

3. A 6.5kW water heater is fed from a domestic 230v supply using PVC/PVC flat twin (thermoplastic 70 degree C). The cable is 20m long and is protected by a BS3036 fuse. The ambient temperature is 35 degrees C and there is no thermal insulation or grouping.
 24 March 2008 03:59 PM
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GaryMo

Posts: 1210
Joined: 09 May 2007

1 - 16mm
2 - 2.5mm
3 - 6mm

Preparing to be shot down!

[EDIT] Keep up the good work. I for one would like to see more examples like this if people have any spare time.

Edited: 24 March 2008 at 04:08 PM by GaryMo
 24 March 2008 05:17 PM
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bakey1959

Posts: 767
Joined: 24 October 2006

Originally posted by: GaryMo

1 - 16mm
2 - 2.5mm
3 - 6mm

[EDIT] Keep up the good work. I for one would like to see more examples like this if people have any spare time.


Spot on 3 out of 3. If you care to write up your working outs and email me, I'll check against mine. As others may want to have a go it is probably too early to post solutions.
 24 March 2008 06:19 PM
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GaryMo

Posts: 1210
Joined: 09 May 2007

Sent you a PM for your email address.
 24 March 2008 07:44 PM
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bakey1959

Posts: 767
Joined: 24 October 2006

Just PM'd back.
 25 March 2008 04:13 PM
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bakey1959

Posts: 767
Joined: 24 October 2006

Hi Gary

Very well done!

Just one minor disagreement...

Questions 2 and 3.

You selected the cable from Table 4D2A, Table 4D5 is the one intended for 'Flat' PVC Sheathed/ PVC Insulated cables (Twin & Earth).

Thanks for pointing out missing installation method in question 2, it would help wouldn't it

Edited: 25 March 2008 at 04:14 PM by bakey1959
 25 March 2008 05:22 PM
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GaryMo

Posts: 1210
Joined: 09 May 2007

Cheers Chris.
I'm a little embarrassed to say but I've always used 4D2A for twin and earth
Not any more though
 25 March 2008 07:55 PM
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bakey1959

Posts: 767
Joined: 24 October 2006

There see, you've now gained an extra amp with your 6mm T&E
 25 March 2008 09:48 PM
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hertzal123

Posts: 330
Joined: 26 August 2007

Bakey,
Is it true,for a fixed load such as heater not subject to overload,factor for 3036 fuse need not be used.Also if all factors do not apply along the whole cable route,ie thermal insulation,do you only apply the factor that gives the highest current carrying capacity.(from Trevor Marks Guide)
Regards,Hz
 25 March 2008 10:22 PM
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bakey1959

Posts: 767
Joined: 24 October 2006

Originally posted by: hertzal123

Bakey,

Is it true,for a fixed load such as heater not subject to overload,factor for 3036 fuse need not be used.Also if all factors do not apply along the whole cable route,ie thermal insulation,do you only apply the factor that gives the highest current carrying capacity.(from Trevor Marks Guide)

Regards,Hz


So here what you would now be referring to in Q3 is 433.3.1 (ii) in BS7671.2008, Yes. So apply the 0.725 factor to socket circuits, lighting circuits, cooker circuits, etc., but not an immersion, ashower, or such fixed loads that "because of the characteristic of the load...is not likely to carry overload current...".
On the second point 523.7 and Table 52.2 gives the derating factor to be applied to the situation.
 26 March 2008 09:54 PM
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AJJewsbury

Posts: 11766
Joined: 13 August 2003

BS7671:2008 requires that: Iz ≥ In ≥ Ib

Strictly speaking, I don't think that's literally true.

In ≥ Ib - yes

Iz ≥ Ib - yes

but Iz ≥ In ≥ Ib would mean that you're requiring Iz ≥ In which is only required if the device is required to protect the cables against overload - sometimes this isn't needed - for example - industrial motor circuits where overload protection is at the motor end, or even the humble unfused spur in a domestic environment - both of which BS 7671 permits.

- Andy.
 27 March 2008 06:11 PM
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bakey1959

Posts: 767
Joined: 24 October 2006

Yes Andy, your right. I should really be using It instead of Iz (as in Appendic 4), and then consider the Iz/In relationship if overload protection is afforded by the device.
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