cg,ca,ci,cc

[lofty84]

a single phase load of a 13a is supplied via a 15 a bs 1361 fuse using single core 70 degree c theromplastic copper cables installed to method 4 (ref method B) the rating factor for grouping is 0.7 and for ambient temperature 0.87 and overload protection is to be provided. the minimum acceptable size cable would be

a)1mm sq
b)1.5mm sq
c)2.5mm sq
d0 4 mm sq

ive used the brb and its told me to go to app 4

I know In is going to be 15


so 15 / ca ci cc

ca 0.87
ci no idea
cc 0.9

this gives me this for an answer but its obviously wrong


15/0.87x0.9 = 19.16


can anyone show me where im going wrong please

 

[ackbarthestar]

There appears to be a reference method of 4 or A which will derate the cable from you maximum clipped direct value. (from your tutor's example )
There will not be Cc (see 433.1.1 ,2 ,3)
You have missed out Cg = 0.7. Since they are all running concurrently I would have thought you include the lot in one formula.

Try
In/(Cg*Ca) then find a suitable sized cable from the correct table and column in appendix 4.

 

[JamesBrownLive]

As your tutor has not mentioned multiple circuits I would not bother with Cg.

I don't think you are wrong with your figure of 19.16, but you seem to be forgetting that the total is an Ampere total not a CSA one. So all you need to do now is to look up which of the cables listed is capable of carrying 19.16A

 

[lofty84]

cheers guys I did try it ur way first ack but it didnt sound right to me I then tried it this way and thought I had the answer but I forgot it was in A i would need to find a cable capable of carring it. the thing is where the bloody hell do I look in the brb for it, im looking from pg 274 onwards but cant find what im after. this is going to take more then 2 mins if i get one like it in the test later

 

[lofty84]

i know the answer is 4mm2 but I want to be able to find and see it with my own eyes

 

[JamesBrownLive]

Doh, sorry the tutor did mention Cg. Overlooked that.

Lets try again...

15A / Ca of 0.87

17.24A / Cg of 0.70

24.63A

Don't think Cc applies in this case as the cable is not buried in the ground (but I'm probably wrong).

Now check Table 4D2A.

 

[lofty84]

nice one guys

 

[ackbarthestar]

Its reasonably easy when you get the hang of it.

Know the type of cable - T&E which could be found in two places, 4D2A or 4D5.

4D2A used to be the table for T&E now flat cables in domestic environments are dealt with in 4D5 and has a reference back to 4D2A.

So column 7 4D5 = 20A min for 2.5mm2, 26A for 4.0mm2 , 32A for 6.0mm2
and Column 2 4D2A = 18.5A for 2.5mm2, 25A for 4.0mm2, 32A for 6.0mm2

In/(Ca*Cg) or 15/0.87 *0.7 = 24.6A

So 4.0mm2 will give overload and short circuit protection .

When do I get my certificate? - BTW Jamesbrownlive doesn't get one since he missed out the grouping factor

 

[lofty84]

An industrial Board supples

- 4 TP motors
- 2 x TP&N motors
- 4 x single phase ring circuits
- 2 x single phase lighting circuits

how many live conductors are there

a)36
b)38
c)40
d)52

anyone know how to work that out ...

 

[ackbarthestar]

First of all where's my cert?

Ask yourself, what does it mean by 'live conductors' ?

Then, usually from experience you will know how many live conductors will go to feed each of the circuits

Here's a start

4 * 3 triple pole 3 phase motors will have 12 conductors
2 * 4 pole 3 phase motors will have 8 conductors
4 * number of live conductors in the pipe or trunking will be?
2 * number of live conductors in the pipe or trunking will be?

Total number of conductors entering the DB will be my invoice in pounds!

 

[telectrix]

- 4 TP motors 4 x 3 =12
- 2 x TP&N motors 2 x 4 = 8
- 4 x single phase ring circuits 4 x 2 x 2 = 16
- 2 x single phase lighting circuits 2 x 2 =4

total 40.

how many live conducto

 

[JamesBrownLive]

When do I get my certificate? - BTW Jamesbrownlive doesn't get one since he missed out the grouping factor

But at least I got the reference method correct... B not A :19:

And does that mean Table 4D2A is still correct as there is no mention of Ref B in Table 4D5 ??

 

[lofty84]

cheers mate is there anywhere in the book i can search for the answer

havent got a clue about this stuff

 

[telectrix]

i'd use table 4d2a as the q did not state flat t/e cable.

 

[lofty84]

I did you no what to times everything by tele

 

[telectrix]

Doh, sorry the tutor did mention Cg. Overlooked that.

Lets try again...

15A / Ca of 0.87

17.24A / Cg of 0.70

24.63A

Don't think Cc applies in this case as the cable is not buried in the ground (but I'm probably wrong).

Now check Table 4D2A.

it's 13A n ot 15, IMO gives 21.35A. so 2.5mm

 

[telectrix]

or, take 2.5mm ref, method C = 27A then multgiply by rating factors. 27 x 0.87 x 0.7 = 16.44, so 2.5mm will handle 13A.

 

[ackbarthestar]

Being wacked with a wet fish excluded, I would say that you maybe right and wrong

Method 4 = A
Method 3 = B

looking at app 4 table 4A2 number 3 = ref method B I think my internal translator got jammed

- I'll just take the money and run

 

[telectrix]

all this brain work is causing alcohol starvation, need beer to focus.

 

[JamesBrownLive]

it's 13A n ot 15, IMO gives 21.35A. so 2.5mm

I was taught that you apply rating factors to the CPD not to the load?? That's certainly how they teach it at www.LearningLounge.com.

Cheers

 

[lofty84]

ur telling me my exams in 2 hrs eeeek


the answer sheet i have gives the answer as 4mm2 so they have used the IN value (15a)


tele how did you no what to times things by on the industrial board question with the motors and that, just hoppe i dont get one of these types but i bet i do ....

 

[lofty84]

1 hr even

 

[ackbarthestar]

or, take 2.5mm ref, method C = 27A then multgiply by rating factors. 27 x 0.87 x 0.7 = 16.44, so 2.5mm will handle 13A.

Probably in real life this would not be appropriate, but I doubt for this example you would use the load to find your tabulated current since your final circuit has not been designed to incorporate overload protection.

Mr Pedant

 

[JamesBrownLive]

Being wacked with a wet fish excluded, I would say that you maybe right and wrong

Method 4 = A
Method 3 = B

looking at app 4 table 4A2 number 3 = ref method B I think my internal translator got jammed

- I'll just take the money and run

Are you looking at the BRB or BGB ??

BRB P.261
Method 3 = A
Method 4 = B

But I agree with Tel.... It's beero'clock.

 

[ackbarthestar]

Are you looking at the BRB or BGB ??

BRB P.261
Method 3 = A
Method 4 = B

But I agree with Tel.... It's beero'clock.

BRB
I was referring to ref methods translated from the 16th ed to 17th ed, not the row numbers in appendix 4 of the 17th ed.

 

[lofty84]

sorry guys but can anyone tell me where in the red book i can look to find out this question An industrial Board supples

- 4 TP motors
- 2 x TP&N motors
- 4 x single phase ring circuits
- 2 x single phase lighting circuits

how many live conductors are there

a)36
b)38
c)40
d)52

anyone know how to work that out ...

 

[JamesBrownLive]

BRB
I was referring to ref methods translated from the 16th ed to 17th ed, not the row numbers in appendix 4 of the 17th ed.

Well then we're both correct :beatnik: now where did I put that beer...

Cheers

 

[ackbarthestar]

sorry guys but can anyone tell me where in the red book i can look to find out this question An industrial Board supples

- 4 TP motors
- 2 x TP&N motors
- 4 x single phase ring circuits
- 2 x single phase lighting circuits

how many live conductors are there

a)36
b)38
c)40
d)52

anyone know how to work that out ...

The answer is knowing what a live conductor is, from part 2 definitions and knowing how many live conductors are required for each type of circuit.

The C&G 2382 level 3 assumes a certain level of installation knowledge before you start.

 

[lofty84]

i did my apprenticeship with a domestic sparky so have no idea unfortuneatly

 

[ackbarthestar]

Well then we're both correct :beatnik: now where did I put that beer...

Cheers

I'll send you a copy of my cert, you can change the name or use mine - up to you.

 

[telectrix]

OK 4 x 3 phase motrors have ( L1.L2,L3 ) x 4 = 12.

2 tp&n HAVE ( L1,L2,L3,N ) x 2 = 8.

4 1 phase RFC hve 2 legs L. 2 legs N = 4 x 4 = 16.

2 1 phase ccts. have 2 x L and 2 x N = 4

as cpc's have not been mentioned then i've ignored them.... total live conductors 40.

 

[ackbarthestar]

i did my apprenticeship with a domestic sparky so have no idea unfortuneatly

Funnily enough much of this stuff is, or should be, taught at college/training centres.

I would find someone who does some industrial/commercial installations where you can link up with to build on your knowledge and experience. However, that doesn't get you though the immediate obstacle.

 

[MarkieSparkie]

a single phase load of a 13a is supplied via a 15 a bs 1361 fuse using single core 70 degree c theromplastic copper cables installed to method 4 (ref method B) the rating factor for grouping is 0.7 and for ambient temperature 0.87 and overload protection is to be provided. the minimum acceptable size cable would be

a)1mm sq
b)1.5mm sq
c)2.5mm sq
d0 4 mm sq

ive used the brb and its told me to go to app 4

I know In is going to be 15


so 15 / ca ci cc

ca 0.87
ci no idea
cc 0.9

this gives me this for an answer but its obviously wrong


15/0.87x0.9 = 19.16


can anyone show me where im going wrong please

Minimum tabulated current required:
It=>In/CaCiCgCc
It=>15A/0.87x1x0.7x1
It=>24.63A
From Table 4D1A for Singles, 70°C, Copper.
It=24A for 2.5mm[SUP]2[/SUP] Does not comply with calculated It.
It=32A for 4mm[SUP]2 [/SUP]selected conductor size, complies with calculated It.

 

[telectrix]

this would be a typical installation of a 3kW immersion heater. who has ever seen one wired in 4mm? i certainly aint.

 

[JamesBrownLive]

Minimum tabulated current required:
It=>In/CaCiCgCc
It=>15A/0.87x1x0.7x1
It=>24.63A
From Table 4D1A for Singles, 70°C, Copper.
It=24A for 2.5mm[SUP]2[/SUP] Does not comply with calculated It.
It=32A for 4mm[SUP]2 [/SUP]selected conductor size, complies with calculated It.

Well done for spotting 'Single Core'. There's me and Ackbar debating which table, and we're both wrong :mad2:

Am forwarding the cert Ackbar forwarded to me, so now it has 2 names crossed out and yours instead :winkiss:

 

[ackbarthestar]

I shall get my hat now - exit stage left :behead::35:

 

[Sparky Ninja]

Looks like a missed out on a nice evening of appendix 4 tosh. Serve's me right for taking mum to Tesco's

:19:

 

[i=p/u]

do you not only use grouping factor for equaly loaded cables and if are 2 times diamter apart then none ata all??

 

[sambotc]

I've just successfully completed my 17th edition paper, and one of the parts which confused me slightly was the correction factors (amongst many other parts not helped by the legal lingo BS7671 is written in) Having read this thread it all seems to make sense now thanks. One thing, presumably being an accumulative calculation, the correction factors have to be applied in the order they are stated e.g CaxCixCgxCcxCt ?

How or where do you apply table 52.2 derating factors, does it replace ci when the length of cable in insulation does not exceed 0.5m? So take a cable clipped direct to timbers in a loft space, which travels 200mm through insulation, would you ignore the ci for clipped direct and use 0.63 in its place?

Hope that makes sense

Cheers, Sam

 

[ackbarthestar]

I've just successfully completed my 17th edition paper, and one of the parts which confused me slightly was the correction factors (amongst many other parts not helped by the legal lingo BS7671 is written in) Having read this thread it all seems to make sense now thanks. One thing, presumably being an accumulative calculation, the correction factors have to be applied in the order they are stated e.g CaxCixCgxCcxCt ?

A I understand it, If the circuits affected by the various self heating effects that would demand correction factors but do not affect the conductors concurrently then apply each correction factor separately and take the worst case (lowest correction factor; highest current value)
If the self heating effect of the environment ie. circuits run together in a group and surrounded by thermal insulation as well as an ambient temperature higher than 30 deg C. then apply all the factors together.
Imagine the example.
10 * 50W downlights on a circuit which is run with 12 other T&E circuits from a distribution board covered in 300mm of thermal insulation
You have 2.2A constant load running in trunking (Method B)
Cg = 0.45
Ci = 0.5
In = 6A
It= 6/(0.45*0.5) = 26.7A or
That would mean you would need to wire the first part of the circuit in 2.5mm2 and then possibly use a smaller conductor csa after that!
If you separated out the self heating conditions then you would only apply the worst case such as It=6/(0.45) = 13.3A.So you would be Ok with 1.0mm2 conductors.


How or where do you apply table 52.2 derating factors, does it replace ci when the length of cable in insulation does not exceed 0.5m? So take a cable clipped direct to timbers in a loft space, which travels 200mm through insulation, would you ignore the ci for clipped direct and use 0.63 in its place?

I would only apply table 52.2 when the cable passes through thermal insulation and not in thermal insulation.

If the cable now only run through 300mm and not in or surrounded by 300mm then you might apply a derating factor of 0.57, but since the grouping factor is lower and not concurrent then you would only apply the worst case which would be the grouping factor of 0.5.


Hope that makes sense

Cheers, Sam

I hope this helps, there are some anomalies but I think you will get the picture