UK Thermal constraints on a ring Electrical design question

[J-Spark]

Hi all,

I am working on my revision for the 2396 exam and wanted to see if someone brainier than me could advise, when completing the thermal constraints equation t=(k²*s²)/I² on a ring circuit do you input the ring for s² as (2*2.5)².

Example1 :
Fault current 211A

t= (115² * 2.5²)/ 211²

Example 2 :
Fault current 211A

t= (115² * {2*2.}²)/ 211²

This doesn't seem to be picked up in any of the books, I'd be grateful for any help or any tips on passing the 2396 this will be my 2nd attempt at the exam.

 

[PS Electrics]

Hi all,

I am working on my revision for the 2396 exam and wanted to see if someone brainier than me could advise, when completing the thermal constraints equation t=(k²*s²)/I² on a ring circuit do you input the ring for s² as (2*2.5)².

Example1 :
Fault current 211A

t= (115² * 2.5²)/ 211²

Example 2 :
Fault current 211A

t= (115² * {2*2.}²)/ 211²

This doesn't seem to be picked up in any of the books, I'd be grateful for any help or any tips on passing the 2396 this will be my 2nd attempt at the exam.

Long time since at college but the calculation is for the limitation of conductor or cable temperature under fault conditions not cables.

 

[s harris]

Hi all,

I am working on my revision for the 2396 exam and wanted to see if someone brainier than me could advise, when completing the thermal constraints equation t=(k²*s²)/I² on a ring circuit do you input the ring for s² as (2*2.5)².

Example1 :
Fault current 211A

t= (115² * 2.5²)/ 211²

Example 2 :
Fault current 211A

t= (115² * {2*2.}²)/ 211²

This doesn't seem to be picked up in any of the books, I'd be grateful for any help or any tips on passing the 2396 this will be my 2nd attempt at the exam.

Right, here goes.

I'm no expert but I reckon its (2.5^2), explanation as below. Would be great to see this answered by others so bumping it up!

For an earth fault current of 211A and you want to see if the 2.5mm cpc is adequate:
the worst case scenario is the current could all go down one leg eg. if there was an undetected break in the ring. So, assuming the OCPD is a 32 Type B, that's confirming your value of t is longer than it takes the ocpd to operate. t= (115² * 2.5²)/ 211², t here meaning the time taken for sustained fault current to take cable to dangerous temperature =1.86s. Longer than 0.1s so ok!

I find it easier to see it as prospective fault energy < energy the cable can healthily withstand: (i^2)t < (s^2)(k^2). And 4452.1 < 82656 so that's ok.

Interestingly a former tutor said that studies had been done on how the current gets distributed in a healthy ring and apparently it's something like a maximum of two thirds of the total current going down one leg, very close that 20A we get told to use on rings instead of In when calculating current carrying capacity.

Oh dear, perhaps you'd better ignore my post as I've confused myself now, imagining a short circuit fault current of 211A dividing into a 1:2 ratio as 70A :141A and ending up heating the cable for 20s (much longer than 1.86s) before the OCPD trips.

I haven't answered your question, just made it worse. Sorry.

 

[s harris]

Correction: Ignore that last bit, I got confused. The OCPD will still trip in time at the sum of the currents.