Adiabatic Calculation

[cprfenom]

When doing the adiabatic calculation for cpc on a circuit greater than 32 amps, do you specify 0.4 or 5 seconds on the trip time?

So my situation would be 40A MCB no RCD 230V and Zs of 0.33 ohms

Kind regards,

Chris

 

[Risteard]

You use the actual time the device will take to operate at the prospective earth fault current at that point.

 

[Rockingit]

^^^ what he said ^^^

 

[Richard Burns]

As Risteard, taking account of any effects of current limiting if applicable.
However for ADS purposes the disconnection time would be 5s, but one would hope that the MCB would operate faster than that!

 

[cprfenom]

As Risteard, taking account of any effects of current limiting if applicable.
However for ADS purposes the disconnection time would be 5s, but one would hope that the MCB would operate faster than that!

Based on 230/0.33 it should trip somewhere between 0.1 and 11 seconds. The issue I have is I don't have the l2t for the breaker to narrow it down. Based on 5 seconds and using k factor of 113 its saying I should have a cpc of about 13.8mm

So I am sat here thinking I am doing something stupid

 

[telectrix]

yes, by using 5 secs. from the Zs, calculate the fault current. then go to app.3 and find out the trip time for that MCB at that fault current. then use the If and the t values in the adiabatic.

 

[cprfenom]

yes, by using 5 secs. from the Zs, calculate the fault current. then go to app.3 and find out the trip time for that MCB at that fault current. then use the If and the t values in the adiabatic.

That's the problem I am having. Based on my readings its 230/0.33=696.96

That is off the scale so is 0.1. If I calculate that then I am getting about 1.95 cpc. Why this bothers me is Im probably wrong or its doesn't sit well with the rule of thumb of the cpc being less than half the phase conductor size as its powering a sub board?

Am I doing this all wrong?

 

[jackhammerJIM]

I get 2mm using a fault current of 696 amps time of 0.1 s and k factor of 113 .

don't have regs to hand to check the time current characteristics

 

[Richard Burns]

For a 40A type B MCB this will trip on the magnetic trip between 3 and 5 times In.
So after 5In (200 A) the magnetic trip will operate and the MCB will switch off in less than 0.001 seconds.

As you have 696 A as the PEFC at that point you can assume that from 200A upwards you will have a trip time of 0.001 seconds.
This will give a minimum csa of 0.2mm² so use an appropriate size that is suitable for the supply.

 

[alanl]

0.001?
That is awfully quick, maybe best to use 0.1?

 

[Richard Burns]

Depends on the circuit breaker if you have manufacturers curves that go down to 0.001 then use that, if the curves stop at 0.01 then use the value of current at that level.
e.g for this curve on type B you may go down to 0.005.

View attachment 60898 tripping curves B C D Eaton.pdf

For this curve then you could go down to 0.001
View attachment BS EN 60898 MCB time current curve Wylex.pdf