17th Edition question


Postby livewire » Wed Jun 17, 2009 7:34 pm

hi

A domestic consumer unit on a TN-C-S system uses a 15A Type 2 mcb, to BS3036 to protect a radial circuit wired in 2.5mm^2 two-core pvc insulated cable with a 1.5mm^2 protective conductor clipped directly to a surface. If this circuit supplies socket outlets for hand held equipment, the maximum length of run is:

a. 35m
b. 43m
c. 54m
d. 58m

The answer from the paper gives a correct answer of 35m.



My assumptions are:

Best way to tackle this problem is to rearrange the formula for voltage drop i.e.

V_drop = [(mV/A/m)*(I_b)*L]/1000

hence

L = [1000*(V_drop)] /[(mV/A/m)*(I_b)]


Problem is the question doesn't give me a value for the design current: I_b.

So I've used a hypothetical value of 14A, since 14A < 15A satisfying I_b < I_n.

Using table 4D5 from the regs I get a voltage drop of 18V (mV/A/m) for installation ref method C clipped direct for 2.5 twin + earth cable.


Using these values in the above formula gives:

L_max = (1000*18)/(18*14) = 71.42m


Some questions then,

Am I missing something?

Am I ok to use a design current of 14A?



help please...
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Postby moggy1968 » Wed Jun 17, 2009 11:36 pm

Google the following: "tlc-direct.co.uk/Technical/Charts/VoltageDrop.html"
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Postby sparx » Thu Jun 18, 2009 12:26 pm

hi live wire, gotta be quick as web site KEEPS timing me out 1/2 way through answer...
basically no such device as BS3036, 15A, type 2.
could be BS3871, 15A, Type 2?
you are doing a long winded calc. (incorrectly in your transposition, using 18 twice instead of 35- clue)
as we are electricians not mathematicians try a table from On site guide, such as 7.1 on page 47.
The only set up in there that seems to fit is for bs 1361 15A FUSE @ 15A
So if you use that data it works out to a Volt drop of just over max allowed at full rated load of 15A which is figure to use, therefor by default it seems you must go to next answer down ie 35M.

I would ask your tutor to show workings as I think info given may be incorrect!
good luck Sparx
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Postby livewire » Thu Jun 18, 2009 3:13 pm

[quote="sparx"]hi live wire, gotta be quick as web site KEEPS timing me out 1/2 way through answer...
basically no such device as BS3036, 15A, type 2.
could be BS3871, 15A, Type 2?
you are doing a long winded calc. (incorrectly in your transposition, using 18 twice instead of 35- clue)
as we are electricians not mathematicians try a table from On site guide, such as 7.1 on page 47.
The only set up in there that seems to fit is for bs 1361 15A FUSE @ 15A
So if you use that data it works out to a Volt drop of just over max allowed at full rated load of 15A which is figure to use, therefor by default it seems you must go to next answer down ie 35M.

I would ask your tutor to show workings as I think info given may be incorrect!
good luck Sparx[/quote]



thanks sparx

I think the whole question is incorrect as it doesn't give enough information to calculate the voltage drop! (i.e. i would need length of cable) in order to work backwards to find the max cable length. if that makes any sense.

The only reason i'm doing it this way is because my tutor has said i can't take the onsite guide into the exam!
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Joined: Wed Apr 15, 2009 4:06 pm


Postby kbrownie » Thu Jun 18, 2009 5:48 pm

[quote="livewire"]
I think the whole question is incorrect as it doesn't give enough information to calculate the voltage drop! (i.e. i would need length of cable) in order to work backwards to find the max cable length. if that makes any sense.

The only reason i'm doing it this way is because my tutor has said i can't take the onsite guide into the exam![/quote]

You do it by working out the per meter voltage drop then multiply until it does not comply to percentage allowed,
What exam is it?
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Postby livewire » Fri Jun 19, 2009 7:27 am

[quote="kbrownie"][quote="livewire"]
I think the whole question is incorrect as it doesn't give enough information to calculate the voltage drop! (i.e. i would need length of cable) in order to work backwards to find the max cable length. if that makes any sense.

The only reason i'm doing it this way is because my tutor has said i can't take the onsite guide into the exam![/quote]

You do it by working out the per meter voltage drop then multiply until it does not comply to percentage allowed,
What exam is it?[/quote]


hi KB

the exam or exams are: city & guilds 2382/92.

I got this question from an example online paper (getting some practice in for the real thing!), I don't think its terribly realistic though!


so KB, you say you can get the same answer as above (using only the regs book) with the information that's given in the question??

It's just that i'm a little confused now!!
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Postby kbrownie » Fri Jun 19, 2009 12:12 pm

you won't need your OSG for 17th eds exam as that is purely going to ask you about the regulation and any calculations that appear are not as complex as the example you have given, they will be very basic. In reality all the answers should be available in the big red book get to know your way round it.
Your exam questions will be in numerical order of the sections in the book although some may cross ref back.
KB
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Postby livewire » Fri Jun 19, 2009 6:29 pm

[quote="kbrownie"]you won't need your OSG for 17th eds exam as that is purely going to ask you about the regulation and any calculations that appear are not as complex as the example you have given, they will be very basic. In reality all the answers should be available in the big red book get to know your way round it.
Your exam questions will be in numerical order of the sections in the book although some may cross ref back.
KB[/quote]

that's good to hear, thanks again KB.
livewire
Posts: 22
Joined: Wed Apr 15, 2009 4:06 pm


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