Bright Sparks

[ricicle]

I got the overall circuit length as 20.24m,
(0.3 x 1000)/(7.41 x 2)

(There was a tiny error in my answers before owing to rounding up/down figures meaning they came out at 20.23m)

Correct - but thats as far as you can go for a correct answer....

To work out:-

A = original length B = new length

New length =R1B+RnB/0.01482
Extra length = (R2B-R1B) / (0.0121-0.00741)

 

[SimonH2]

You conduct a resistance test on the TOTAL circuit:

R1+R2 = 0.4
R1+Rn = 0.3

Hmm, guesswork here.

Use this data for your calculation

1.5mm = 12.1
2.5mm = 7.41

More guesswork.

0.4 of what ? Linguinis ? Inches of damp string ?
12.1 linguini / Blackpool tram ?

I don't think it is possible, maybe the OP can enlighten us with the working out?
Unless I am allowed a negative answer

-1.09m for the initial circuit
21.32m for the addition

I got the overall circuit length as 20.24m,
(0.3 x 1000)/(7.41 x 2)

(There was a tiny error in my answers before owing to rounding up/down figures meaning they came out at 20.23m)

Funnily enough, exactly the figures I get. That's if we make assumptions about the figures given - like assuming that the first figures are ohms, and the second figures are milliohms/metre of conductor - they differ from the figures in BS7671 which gives from which the values can be derived as 14.5 and 9.

 

[Spark123]

12.1mΩ/metre and 7.41mΩ/metre are the resistance figures for 1.5mm² and 2.5mm² conductors at 20°C from the appendix in GN3 and iirc are also in the on site guide.
BS7671 tends to use 70°C or 90°C figures hence the reason they are higher.

[code:1]
Overall length

7.41 x 2 (2.5mm) conductors = 14.82milliohms/metre

0.3 x 1000 = 14.82t

(0.3 x 1000) / 14.82 = t = 20.24metres

Original leg: (a)

1 x 1.5mm conductor and 1 x 2.5mm conductor = 19.51milliohms/metre

0.4 x 1000 = 14.82a + 19.51b

20.24 = a + b
20.24-a = b

400 = 14.82a + 19.51(20.24-a)

400 = 14.82a + 394.88 - 19.51a

5.12 = -4.69a

5.12/-4.69 = a = -1.09m

Extended leg (b)
(could just work it out using 20.24 + 1.09 but no fun in that)

20.24 - b = a

0.4 x 1000 = 14.82a + 19.51b

400 = 14.82 (20.24-b) + 19.51b

400 = 300 - 14.82b + 19.51b

100 = 4.69b

100/4.69 = b = 21.32m
[/code:1]

 

[EFLImpudence]

I think dizz has pulled a fast one - on purpose or not we do not know.

You do not need to work out any lengths.

The additional 0.1Ω is the extra resistance in the 1.5mm² cpc compared to the 2.5mm² L or N.
In order for this to be the case R1 must be 0.15Ω & R2 (R1 x 1&#8532 will be 0.25Ω

This adds up to 0.4Ω - the total of the circuit - where, indeed, R1 + Rn is 0.3Ω.



Out of interest this works out at 20.24 or 20.5 metres (because his resistance figures do not match).
Obviously, the cable with the 2.5mm² cpc has not been used.

 

[Spark123]

I took it to mean the CPC was 2.5mm and extended with 1.5mm?

 

[EFLImpudence]

I took it to mean the CPC was 2.5mm and extended with 1.5mm?

It did but there isn't room.
As I said the 2.5/1.5 'uses' all the resistance.

 

[ban-all-sheds]

 

[ebee]

The answer is

"A Lemon"

 

[ebee]

My old Grandad used to ask a couple of questions that we considered to be "Old Chestnuts".

Solveable thru trial & error but also by maths.

1/
Two farmers meet at crossroads and get talking.
Each have a herd of sheep.
One says
"Sell me two of yours and I will have twice as many as you"
Tother says
"Sell me two of yours and we`ll each have the same"
How many sheep had each farmer?

2/
(Best spoken not written but here goes!)
If a brick weighs 7 pounds and half a brick then what does a brick and a half weigh?

To the OP,

Better still, work out on the double helix R1 + R2 test the difference in readings around the ring (Crossing over the Ph & cpc) answer 3.25% (Whoops sorry =- make that 6.25% - memory getting old)

 

[dizz]

A big FAT nothing. Looks like you had a go at posting something, but decided to delete it - after all, if your calculation is WRONG for the length of cable, that could be DANGEROUS.

So, just because you are familiar with EIC forms, and know how to complete them that makes you a compentent spark?

So, I chose to give you something original to calculate, but because you cannot look it up, or churn it out, or use industry specific acronyms, you fall short - you've been found wanting - So invite BAS round to do some sparks work, and you'll be OK as long as it's in the book, but anything that involves a bit of maths and you may find him wanting.

 

[JohnW2]

The additional 0.1Ω is the extra resistance in the 1.5mm² cpc compared to the 2.5mm² L or N.

Indeed.

In order for this to be the case R1 must be 0.15Ω & R2 (R1 x 1&#8532 will be 0.25Ω

No, I don't think so! R2 = (R1 x 1&#8532 would only be true if the entire length of CPC were 1.5mm² - but, in fact, it's part 1.5mm² and part 2.5²mm. In other words, in the actual situation:

• R2 = R1 x (A + {1⅔ x B})

... where A is the fraction of total cable lengthg which has 2.5mm² CPC and B is the fraction which has 1.5mm² CPC. As you actually go on to effectively admit, your expression would only become equivalent to the correct one if A were zero (i.e. none of the cable with 2.5mm² CPC were used).

KInd Regards, John.

 

[Spark123]

A big FAT nothing. Looks like you had a go at posting something, but decided to delete it - after all, if your calculation is WRONG for the length of cable, that could be DANGEROUS.

So, just because you are familiar with EIC forms, and know how to complete them that makes you a compentent spark?

So, I chose to give you something original to calculate, but because you cannot look it up, or churn it out, or use industry specific acronyms, you fall short - you've been found wanting - So invite BAS round to do some sparks work, and you'll be OK as long as it's in the book, but anything that involves a bit of maths and you may find him wanting.

I don't think BAS is actually a spark, and anyway - what is the answer to the question as I ended up with a -ve one and that can't be right, or was it just something you made up to try and prove a point but has backfired?

 

[ebee]

Assuming a radial circuit (you did not say otherwise!).
The figure 0.3 ohms gives a length of 20.2491 metres for two conductors (R1 + Rn) of 7.41/1000 ohms per metre.
The figure 0.4 ohms gives a length of 20.4918 metres for two conductors (R1 + R2) 12.1/1000 + 7.41/1000 = 19.52/1000 ohms per metre.

So your figures do not tally unless we accept that they are very approx.

If we accept that they are very approx then some figure like almost zero original cable with all three conductors of 2.5 mm sq csa added to just over 20 metres of cable 2.5/1.5 cable will just about fit into the 0.4 & 0.3 measured resistances .

PS Did our Mr Sheds claim to be an Electrician/Sparky or is he a Non Electrician/Sparky?

 

[SimonH2]

PS Did our Mr Sheds claim to be an Electrician/Sparky or is he a Non Electrician/Sparky?

No, but he does presume to lecture others about something he is apparently neither experienced in or qualified to do. He is very happy to quote regs till the cows come home, and happy to accuse others of being dangerous if they can't recite them without getting a comma in the wrong place, but apparently doesn't actually know all that much about real world electrics (he demonstrated that much in one of my recent threads).

You would most definitely not want him doing any electrics in the real world.

 

[chapeau]

He is the only person in the past twenty or so years on the whole internet I have ever put on ignore. When he comes along and abuses me now I cannot read it, therefore he doesn't get a reply, must drive him bananas.