Disconnect times for adiabatic

[Steviet]

Am I right In assuming that as long as you have worked out the fault current from a zr that meets regs for the breaker you can just use the disconnection time of 0.4 etc instead of using the graphs which I can never seem to understand

 

[Steviet]

I meant zs obviously instead of zr

 

[Simonslimline]

Use the time it takes the ipf to operate the protective device.

Take this info from the graphs in BS7671 or from manufactures data. What dont you understand about the graphs?

 

[Steviet]

For example if I've got 32a bs60898 type by on a circuit of 0.55 zs this gives me 418A I can't seem to link this up to a disconnection time on the graphs I just don't get it aha

 

[Silly Sausage]

They are logarithmic graphs, that's why the scales are non-linear.
From left to right, starting at 1, the increments are 1 up to 10, then increments of 10 up to 100, then increments of 100 up to 1000, etc etc ...
For 418A, call it 400A, look at the 3rd line after 100, that's 400A.
Then look where the curve cuts that line, then the value on the time axis is your disconnection time.

BUT, a 32A curve is all to the left of the 400A line, so take the lowest time on the 32A curve as your disconnection time, that is 0.1sec


Hope that helps, but it's possibly confused you even more.
Blame my chronic explanation!!!

 

[Steviet]

I think you have just made it make sense if my fault current doesn't match up to the curve of the breaker and its higher it'll be the lowest value of that curve and obviously it matches the curve it will then match up to a time higher than the lowest value. Now I'm sounding confusing but I think I understand it aha