Determining the cas of the CPC

[lofty84]

A circuit is protected by a BS88 fuse and the prospective fault current is 171 A. How do I determine the csa of the CPC assuming 70 oC themoplastic cables with copper conductors.

Can anyone tell me where I need to be looking in the BS 7671 or onsite guide please ?

 

[GrannySmiff]

to work out the size of cpc needed you need to understand the adiabatic equation, look that up.

Here is a thread about it, not sure if its any good as only scanned through it.

http://www.electriciansforums.net/i...al-installations/5179-adiabatic-equation.html

 

[pushrod]

Have a look at reg 543.1.3 formula and explanation is there.

 

[i=p/u]

equation
aint in on-site-guide

 

[j.mawer]

Was there any more information in the question? What about for instance the current rating of the fuse? pg 247-248 will give you the time for the fuse to blow which would be required to use the adiabatic equation to get an optimal CSA

 

[lofty84]

no mate thats all it says.

adiabatic equation = S = the sq root of Isq x t / k

s = minimum size of earthing conductor (CPC)?

= square root of 171 sq x 0.1 / 115

= 0.47mm2

Ive no idea if that is right but I have had a go, what has confussed me is the value to put in for t. Im not sure why I chose 0.1 but I did, can anyone say where Ive gone wrong ??

 

[pushrod]

That is correct, if you are correct that the fuse is disconnecting the fault in 0.1 seconds.

Usually you would have to look up the disconnection time in appendix 3 as J.mawer said above. Sometimes they give you an assumed disconnection time in the question. 0.1 S is a common value to use.

 

[ezzzekiel]

someone posted an excellent worked example maybe look in the wiki it prob there.

 

[lofty84]

cheers guys, can I ask more questions ?

 

[pushrod]

Only if you hit the "thanks" button

 

[lofty84]

does anyone know why the adiabatic equation may be used in preference to the selection tables to determine the size of an earthing conductor ?

 

[pushrod]

The adiabatic equation will usually give a lower answer and so if you calculate it you may find that the protective conductor does not need upgrading when a table might suggest that it does.

 

[lofty84]

A water heater with an output of 5kw is connected to a 230v supply via a bs 60898 type b protective device. the cable is 70 oC PVC insulated thermoplastic coper conductors and the length of cable is 18m. the ambient temperature is 35oC. The cable runs for 300 mm through thermal insualtion. it runs together with one other cable for most of its length and is clipped direct to the wall. Determine the following and show all calcs and refs to tables.

1. Design Current

Ib = 5000/230 = 21.73 A

No diversity according to the on site guide (page 97)

Minimum protective device rateing

In is greater then or equal to Ib = 25 A mcb BS 60898

Correction Factors

It ? = In/caxcgxcixcc

ca=table 6a OSG/ cg = table 6c OSG/ ci = i think table 6b OSG/ cc = 1 i think page 124 OSG ??

It = 25 / 0.94 x 0.80 x 0.51 x 1 = 65.18 A

not sure what do do or say for the minimum cable rating ??

Minimum cable size

= 10mm2

Maximum allowable voltage drop

= 4.4 according to table 6D on page 130 OSG

Actual Voltage Drop

= (mv/a/m) x ib x l / 1000

= 4.4 x 21.73 x 18 / 1000

= 1.72 A

is it 5 % of 230 (11.5) so yes that means it doesnt need a bigger cable ?

 

[pushrod]

A water heater with an output of 5kw is connected to a 230v supply via a bs 60898 type b protective device. the cable is 70 oC PVC insulated thermoplastic coper conductors and the length of cable is 18m. the ambient temperature is 35oC. The cable runs for 300 mm through thermal insualtion. it runs together with one other cable for most of its length and is clipped direct to the wall. Determine the following and show all calcs and refs to tables.

1. Design Current

Ib = 5000/230 = 21.73 A

No diversity according to the on site guide (page 97)

correct
Minimum protective device rateing

In is greater then or equal to Ib = 25 A mcb BS 60898
correct
Correction Factors

It ? = In/caxcgxcixcc

ca=table 6a OSG/ cg = table 6c OSG/ ci = i think table 6b OSG/ cc = 1 i think page 124 OSG ??
correct selection of factors and use of tables

It = 25 / 0.94 x 0.80 x 0.51 x 1 = 65.18 A
correct maths but see below
not sure what do do or say for the minimum cable rating ??

Minimum cable size

= 10mm2 According to table 6F (osg) and your result you would need 16mm

Maximum allowable voltage drop

= 4.4 according to table 6D on page 130 OSG

Actual Voltage Drop

= (mv/a/m) x ib x l / 1000

= 4.4 x 21.73 x 18 / 1000

= 1.72 A Volts not amps

is it 5 % of 230 (11.5) so yes that means it doesnt need a bigger cable ?

What you have done is pretty good but not necessarily completely correct
because argueably the nature of the load and the rating factors need to be simultaneously occurring - give me a bit of time and i'll explain.

 

[pushrod]

Because the heater is purely resistive and not subject to an overload, like a motor, when you calc It instead of using the In figure of 25A you can use the Ib figure of 21.73A (your tutor might not want you to but that is how i was taught)

Also the factors have to be simultaneously occurring - from the question it looks to me as if they are not and so you just go with the worst case scenario of the Ci (insulation factor) which would bring you right down to an It=42.6 then in table 6F, 6mm can go up to 47 so 6mm would be ok.

Then your Vd factor would be 7.3mV/A/m. which would now , i think, give a Vd of 2.86V - still ok.
hope that helps, good on you for doing it yourself - that is the way to learn.