MCB tripping time equation?

[pushrod]

Evening,

Can someone tell me what the equation is to work out how quick a CB will trip from the Amps its getting?
I know there are tables in 7671, but I'm sure there is a formula where you can work it out from the fault current amps etc?

Thanks in advance
Craig

Are you thinking of what current will trip an mcb in 0.1 to 5s?

If so, as i=p/u said, for a B type it is 5x In, for a C type it is 10x and for a D it is 20x.
So a 6 amp C type trips in 0.1 to 5s if a 60amp fault current flows.
For a 6amp type D it would be 120 amps

edit; oops hadn't seen 2nd page where tony had also expanded on i=p/u's post .

 

[Engineer54]

The time/current curves gives you all you need, in an easy to read graph....

 

[pushrod]

The time/current curves gives you all you need, in an easy to read graph....

I don't know that i would describe logarithmic scales as easy to read - in my experience people need quite a bit of practice the first time they encounter them

 

[Engineer54]

I don't know that i would describe logarithmic scales as easy to read - in my experience people need quite a bit of practice the first time they encounter them

I take it they do still teach trainee's to read time/current curves?? But yes maybe the first few times is a little daunting, but i'm sure that a qualified electrician would have no problem reading and understanding these graphs...

 

[CRAIGNEWHAM]

Maybe I'm wrong then about the equation.
I'm fine with the graph anyway
Thanks for the replies

 

[telectrix]

what's a graph? ( joking of course )

 

[Chr!s]

Well I'm ok with tha graph etc, I just thought there was an equation where if you had a fault current you could work out how quick it would trip?
I just want to find it for my peace of mind! been looking for hours, but if there isnt one then my mistake!

If all your trying too find is disconnection times you been given the Method.

example

6 amp type B, so 5 x 6 = 30 amps, so as long as you have more than 30 amps you know it will disconnect at least with .4 of a second.

 

[Martin Bradley]

eg voltage divided by say if 60898 type B (5 times rating of braker eg 32A)
so 230/(5*32)= 1.44 ohms
so 230/(5*1.44)= 32A
Type B 5x, C 10x, D 20x.

Also as Chr!s says.

any good for you

 

[Engineer54]

Isn't this information also given in the time/current curves of each type of protective device in an adjacent block, showing MCB/fuse ratings and the required current to achieve disconnection times?? I'm sure it was in the 16th edition...

 

[CRAIGNEWHAM]

Thanks for the replies.
I'm ok with B,C type and their tripping times/currents etc, I just thought there was a formula where by you had a fault current and a CB rating, and you could calculate the CB's tripping time.
Not to worry

 

[MKS]

are you looking for t=k²S²/I² I could be completely wrong, so you could ignore it..

 

[Engineer54]

Well, you could use the easily read time/current curve graphs, that are standard for each type of MCB breakers...