BS88 max ZS table for fuses over 63A TT or not TT

[John-]

I think i got to the same / similar answer:

r1 will always increase as you move away from the supply.
A will also increase as you move away from the supply
B will decrease as you move away from the supply
C will be a constant but is in series with B.
Both B+C are in parallel with A.

Therefore

r1+r2 = r1+ 1/((1/A)+(1/(B+C)))



Bugger, wont let me attach a spreadsheet.



But yea, around 50m i make it

 

[pc1966]

I think i got to the same / similar answer:

r1 will always increase as you move away from the supply.
A will also increase as you move away from the supply
B will decrease as you move away from the supply
C will be a constant but is in series with B.
Both B+C are in parallel with A.

Therefore

r1+r2 = r1+ 1/((1/A)+(1/(B+C)))

Basically that is how I calculate it, but using some python script so I get a pretty graph out of it, etc. Putting in your numbers I get:

Enter line R1 in mOhm/m 0.55
Enter CPC R2 in mOhm/m 0.55
Enter armour R2 in mOhm/m 1.6
Enter length in meters 107
Maximum fault R1+R2 = 0.104 at 96.7 meters



This forum won't allow spreadsheets, no doubt it won't allow python though I could just paste the text to allow others to use it / check it.

 

[pc1966]

Realistically you will struggle to meet disconnection for 107m under those values on the OCPD alone.

You might find it better to look at putting in a fancy adjustable MCCB style of RCD at the feed end set for 1 sec delay and 0.5A or more trip so you might get some selectivity with final circuit 50A BS88 fuses if they can clear a fault in under 1s (as well as no issue with double pole 30mA RCBO or RCD)

You might even get away then with 4-core and parallel 10mm together with the armour as CPC, saving a small bit to help cover the fancy RCD cost!

 

[pc1966]

Another option would be parallel runs of 16mm SWA or similar for each pump if that would allow 50A switch-fuses at supply to disconnect, but location for isolation might require another one closer, etc.

 

[pc1966]

Two 16mm 4-core in parallel have enough SWA to meet TN-C-S 10mm copper CPC equivalent and if putting in values it comes to:

Enter line R1 in mOhm/m 1.15
Enter CPC R2 in mOhm/m 3.1
Enter armour R2 in mOhm/m 3.1
Enter length in meters 107
Maximum fault R1+R2 = 0.289 at 107.0 meters

That along with your supply Ze would be fine for 50A fuse on 5s disconnection (0.27 + 0.29 = 0.56 < 0.79 ohm). You would need yet another SWA for the ancillary stuff to a small CU but if 10-20A then it could be smaller, even single-phase, or maybe just more 16mm.

Cost would need looking in to, but it might actually be easier to wrangle 3 * 16mm 4-core in one duct/route than 35mm 5 core.

 

[John-]

Another option would be parallel runs of 16mm SWA or similar for each pump if that would allow 50A switch-fuses at supply to disconnect, but location for isolation might require another one closer, etc.

Overloads are already provided in the controller.
The run should not be anywhere near 100m...
More worryingly for me right now is that our answers do not match, inclined to think that it is my maths not yours .

I would love to have a look at your code, but right now i need to get my head around what i am doing wrong here...

I found some bracketing errors and sorted those but still not matching. Then i realised that i may not have solved the parallel resistance values correctly for A, B and C?

I am now wondering my assumption that B and C are a simple series circuit in parallel with A is the issue:

This is the formula i used

r1 + (1/((1/(B + C)) + (1/A))) = r1+r2

r1 is in series with A, B and C so no special care needed for r1.

A has a parallel path back with B and C. And B and C are in series, but collectively in parallel with A.

Any ideas please?



This is the spreadsheet with the formulas revealed-: -


Cheers

John

 

[pc1966]

The effective R2 is A in parallel with B+C as you say, can look later at your numbers and another look at my code as well, out at the site so don't have it with me!

 

[John-]

Thank you , that would be great .

I didn't know if it was more complex, along the lines with this example - but it seem you agree with me and therefore it is not: -



If my math is correct i am scuppered anyway, because at just after 40m 5 core 35mm i exceed max ZS just after 40m away form the supply: -

 

[John-]

Spotted an error on the value for R1 for 35mm, i incorrectly stated r1+r2 = 1.1, but is should be 1.048 - not that it made any significant difference.

 

[pc1966]

Checking the numbers I used before getting 0.104 ohm at 96.7m of 107m we have:
R1 = 0.55 * 96.7 / 1000 = 0.053185
A = 1.6 * 96.7 / 1000 = 0.15472
B = 1.6 * (107 - 96.7) / 1000 = 0.01648
C = 0.55 * 107 / 1000 = 0.05885
B+C = 0.07533
R2 = A||(B+C) = 0.050663
R1+R2 = 0.103848 which is 0.104 ohm to 3 digits.

 

[pc1966]

Overloads are already provided in the controller.

It is for fault protection, and to prevent a fault taking out the DNO fuse (which is basically the same as protecting your own 100A fused-switch).

The run should not be anywhere near 100m...
More worryingly for me right now is that our answers do not match, inclined to think that it is my maths not yours .

Try it with some of the numbers I just did to see if they agree at all, I can post my script later as text if anyone wants to play with it. In the mean time I re-ran a manual calculation for the solution my script found and it seems to make sense.

 

[pc1966]

OK, got my script and edited to add end-of-cable result, ran with your values:
Enter line R1 in mOhm/m .55
Enter CPC R2 in mOhm/m .55
Enter armour R2 in mOhm/m 1.6
Enter length in meters 100
Maximum fault R1+R2 = 0.097 at 90.4 meters
End fault R1+R2 = 0.096 at 100.0 meters

 

[John-]

OK, got my script and edited to add end-of-cable result, ran with your values:
Enter line R1 in mOhm/m .55
Enter CPC R2 in mOhm/m .55
Enter armour R2 in mOhm/m 1.6
Enter length in meters 100
Maximum fault R1+R2 = 0.097 at 90.4 meters
End fault R1+R2 = 0.096 at 100.0 m