Adiabatic Equation

[Ben Murray]

Hi All,

I think I have confused myself somehow and feel I am spiraling into further confusion. When I am doing the equation, it is saying the longer the cable is, the smaller the CPC has to be. My calculations are below.

20A radial BS61009-1 Curve C
Install method type B using 2.5 singles
Ambient 30c
Volt drop is fine (1%)

OSG table I1 = 14.82
OSG table I2 = 1.04
Length = 6.3m
14.82 x 1.04 x 6.3 / 1000 = 0.10
+ Ze 0.13 = Design Zs of 0.23

230v/0.23 Zs = 962.1A
Tripping time - 0.1s

962.1 x 962.1 = 925,636.41 x 0.1 = 92,563.64
Sqr Rt 92,563.64 = 304.24
304.24 / 115 (k) = 2.65mm

I won't show the working out but if I increase the cable length to 20m, my adiabatic answer becomes 1.37mm

I know I wouldn't design a 20A Radial using 4mm cable but I can't seem to pull myself out of this hole.

Thank you in advance for your help.

 

[MarkyMark]

A 20A type C breaker will trip out at 5 - 10 X In, which is 200A Using this figure give you a size of 0.54mm

 

[andyb]

Don't forget Ben that when you increase the length of the circuit you increase Zs, therefore the fault current will be less.

 

[MarkyMark]

With regards to wiring a 20A circuit using 4mm this is perfectly feasible dependant on cable types and correction factors, i.e grouping, ambient temperature etc

 

[Ben Murray]

A 20A type C breaker will trip out at 5 - 10 X In, which is 200A Using this figure give you a size of 0.54mm

Thank you for your reply.
So to confirm, when working this out, as soon as the fault current reaches the point the breaker will trip in 0.1s (as you said, 200A above, or 100A if it was a type B...) I should always use that A, not the calculated V ÷ R = I.

 

[Strima]

Are you referencing the trip curves in BS7671 to get your disconnection time?

 

[Ian1981]

The time will be less than 0.1 seconds given in appendix 3 so you require the I2t value let through energy of the protective device being less than or equal to the energy withstand of the conductors (K2S2)
Regulation 434.5.2.
For circuit breakers the cpc CSA can be surprisingly lower than you’d think

 

[Pretty Mouth]

What Ian said.

Use the fault current at the origin of the circuit, ie the PEFC. If you use manufacturer's I²t data, you'll get a much smaller permissible cpc. Here's data for hager circuit breakers.
[automerge]1571257850[/automerge]
Hager's curves, if you're interested:

 

[Ben Murray]

What Ian said.

Use the fault current at the origin of the circuit, ie the PEFC. If you use manufacturer's I²t data, you'll get a much smaller permissible cpc. Here's data for hager circuit breakers.
[automerge]1571257850[/automerge]
Hager's curves, if you're interested:

Thank you, I have designed this using Hager, so I will look at these tomorrow.
[automerge]1571259459[/automerge]

The time will be less than 0.1 seconds given in appendix 3 so you require the I2t value let through energy of the protective device being less than or equal to the energy withstand of the conductors (K2S2)
Regulation 434.5.2.
For circuit breakers the cpc CSA can be surprisingly lower than you’d think

Thank you, the disconnection time I am happy with, it was the confusion with the equation coming up with a CPC size of 4mm for a 6m long 20A radial. But I think my question has been answered, much appreciated all. Now onto what feels like my 100th hour for the 2396 design.

 

[Strima]

Now onto what feels like my 100th hour for the 2396 design.

When you get to what feels like your 1000th hour you might be somewhere near... ?

 

[Ben Murray]

When you get to what feels like your 1000th hour you might be somewhere near... ?

HAHA! For something C&G suggest will take 40 hours! My employer is certainly starting to notice my absence lol

 

[HappyHippyDad]

Hi All,

I think I have confused myself somehow and feel I am spiraling into further confusion. When I am doing the equation, it is saying the longer the cable is, the smaller the CPC has to be. My calculations are below.

20A radial BS61009-1 Curve C
Install method type B using 2.5 singles
Ambient 30c
Volt drop is fine (1%)

OSG table I1 = 14.82
OSG table I2 = 1.04
Length = 6.3m
14.82 x 1.04 x 6.3 / 1000 = 0.10
+ Ze 0.13 = Design Zs of 0.23

230v/0.23 Zs = 962.1A
Tripping time - 0.1s

962.1 x 962.1 = 925,636.41 x 0.1 = 92,563.64
Sqr Rt 92,563.64 = 304.24
304.24 / 115 (k) = 2.65mm

I won't show the working out but if I increase the cable length to 20m, my adiabatic answer becomes 1.37mm

I know I wouldn't design a 20A Radial using 4mm cable but I can't seem to pull myself out of this hole.

Thank you in advance for your help.

I haven't read the whole thread but I think someone mentioned getting the correct value for t.

You have chosen to use the time current characteristics of Appendix 2 which will only give a t = 0.1 when the line becomes vertical on the graph. You need more accurate figures for your t.

If you think about it in your second example of 20m, the Zs increases to 0.44 so in reality the value of t will be greater for this than the value of t would be for your Zs of 0.23 (with regards the 6.3m run). This greater value of t would then affect the value of S in the adiabatic and give you a higher figure for the 20m run compared to the 6.3m run.

Check your maths for your Zs for the 6.3m run. You have 230/0.23 = 926.1A, whereas I make it 1000A.

 

[Ben Murray]

I haven't read the whole thread but I think someone mentioned getting the correct value for t.

You have chosen to use the time current characteristics of Appendix 2 which will only give a t = 0.1 when the line becomes vertical on the graph. You need more accurate figures for your t.

If you think about it in your second example of 20m, the Zs increases to 0.44 so in reality the value of t will be greater for this than the value of t would be for your Zs of 0.23 (with regards the 6.3m run). This greater value of t would then affect the value of S in the adiabatic and give you a higher figure for the 20m run compared to the 6.3m run.

Check your maths for your Zs for the 6.3m run. You have 230/0.23 = 926.1A, whereas I make it 1000A.

I forgot to type my correction factor of Cmin - 0.95, this is where it drops to my answer.

Good point about the Zs increase, like I mentioned, I was flustered yesterday and couldn't see past the spreadsheet of numbers!

I have changed a few equations and numbers and I am now back on the straight and narrow.