where to find the protective device operating current (I2) in 7671???

[LukeScotty]

does anyone know where in the bs:7671, you can find the protective device operating current (I2) ? Ive been looking for quite some time now, and i'm getting rather fustrated, because I cant find it?

cheers luke

 

[alan]

I think you mean the Time/current graphs starting on page 194-201

 

[TonyM58]

Luke, you can look at this a few diferent ways. As Alan said, you can look at the graphs in the regs, or if you want the instanteneous trip current for cicrcuit breaker type, go to table 7.2b, page 51B

it depends how you phrase the question:-

If you said, how much current does it take for a 32A Type B BS 60898 circuit breaker to operate in a given time (say 0.4S), then you would be on page 199 in the regs. (or how long will it take to operate with a set/calculated/given fault current)

Wheras if you wanted to know the 'trip current' for the same MCB, you would be in table 7.2B This would tell you that a type b has an instantaneous trip current of 3 to 5 times In, so it would take 90-150A to trip the MCB

so i suppose it depends if you are foccussing on time are actual current levels.

To be technically correct, a DEVICE does not have an I2, a CIRCUIT has an I2

I2 is the 'current causing effective operation of the protective device', therefore the I2 value will be dependent on the circuit charecteristics (Zs etc), this will be co-ordinated with the In of the protective device but the two are not the same.

Have a look at 433-02-01 which may help! (or not!)

regards

Tony

 

[LukeScotty]

I'm just trying to understand this statement

"the current (I2) causing effective operation of the protective device devices does not exceed 1.45 times the lowest of the current-carrying capacities (Iz) of any of the conductors of the circuit"

I.E
a 1.5 cable has the current carrying capacity of 20A if Clipped direct, but take into account the lowest? correct? which is (14.5A)

1.45A x 14.5A = 21.025A

so, causing effective operation of the protective device devices does not exceed 21.025A what do I need to look at next ? type of system (TN) and a maximum disconnection time of a TN system is 0.4's (230v) and now i'm lost if i've been going in the right direction

cheers lukey

 

[alan]

You start at the designer current Ib which has to be less than fuse setting In
So you find the nearest fuse or mcb cr =1 if bs3036 cr=0.725
so cable current you chose has to be higher or equal to Ib/craig
cr = fuse fusing factor a= ambient temp i = insulation g = grouping factor
Then when you worked that out find the current carry cable page 220 onwards
OR if it is a convention type circuit looked at the OSG TABLE 7.1 PAGE 42 ONWARDS it
is all worked out for you. which takes into account Zs for 0.4 and 5secs.

 

[TonyM58]

Luke,

we are going deep on this one, its difficult because i dont know the full circuimsatances (and context) of the question

I will try and expand on what i have already said

Your calculations are a bit off

"the current (I2) causing effective operation of the protective device devices does not exceed 1.45 times the lowest of the current-carrying capacities (Iz) of any of the conductors of the circuit"

My interpretation: if you were using singles (for example) and worked out your Iz such that you needed 1.5mm conductors, and you happened to have 2.5mm phase and 1.5mm neutral, then your calcs would be based on the 1.5mm as the "lowest of the current carrying capacities"

Lets say lighting circuit. Ib = 3A (for example)

MCB = 6A so In greater then/ equal to Ib.

Iz for tw/e method 1 =16A

So Ib In Iz is satisfied

Now look at the statement
"the current (I2) causing effective operation of the protective device devices does not exceed 1.45 times the lowest of the current-carrying capacities (Iz) of any of the conductors of the circuit"

So 1.45 x Iz = 1.45 x 16 = 23.2A

So the current causing operation of the MCB should not exceed 23.2A

Type B 60898 has instanteneuos trip current of 3 to 5 x In = 3 to 5 x 6A = 18 - 30A

18 is less than 23.2 so its ok.

But it doesnt matter because :

433-02-01

(ii) its nominal current or current setting (In) does not exceed the lowest of the current carrying capacities (Iz) of any conductors of the circuit, and

(iii) the current (I2) causing effective operation of the protective device devices does not exceed 1.45 times the lowest of the current-carrying capacities (Iz) of any of the conductors of the circuit"

and 433-02-02 says

"where the device is a fuse.....60898......etc compliance with (ii) also results in compliance with condition (iii)

So basically dude they have worked it out for you! This is why this is a grey area which people tend not to refer to: if you have followed Ib In Iz and you are using a standard protective device then I2 (and so 433-02-01 (iii)) doesnt really ever come into it!

You are either now happy

or in a worse state of confusion than when you started

and my typing hand is hurting

have fun!!!

 

[Keith R]

Luke,

we are going deep on this one, its difficult because i dont know the full circuimsatances (and context) of the question

I will try and expand on what i have already said

Your calculations are a bit off

"the current (I2) causing effective operation of the protective device devices does not exceed 1.45 times the lowest of the current-carrying capacities (Iz) of any of the conductors of the circuit"

My interpretation: if you were using singles (for example) and worked out your Iz such that you needed 1.5mm conductors, and you happened to have 2.5mm phase and 1.5mm neutral, then your calcs would be based on the 1.5mm as the "lowest of the current carrying capacities"

Lets say lighting circuit. Ib = 3A (for example)

MCB = 6A so In greater then/ equal to Ib.

Iz for tw/e method 1 =16A

So Ib In Iz is satisfied

Now look at the statement
"the current (I2) causing effective operation of the protective device devices does not exceed 1.45 times the lowest of the current-carrying capacities (Iz) of any of the conductors of the circuit"

So 1.45 x Iz = 1.45 x 16 = 23.2A

So the current causing operation of the MCB should not exceed 23.2A

Type B 60898 has instanteneuos trip current of 3 to 5 x In = 3 to 5 x 6A = 18 - 30A

18 is less than 23.2 so its ok.

But it doesnt matter because :

433-02-01

(ii) its nominal current or current setting (In) does not exceed the lowest of the current carrying capacities (Iz) of any conductors of the circuit, and

(iii) the current (I2) causing effective operation of the protective device devices does not exceed 1.45 times the lowest of the current-carrying capacities (Iz) of any of the conductors of the circuit"

and 433-02-02 says

"where the device is a fuse.....60898......etc compliance with (ii) also results in compliance with condition (iii)

So basically dude they have worked it out for you! This is why this is a grey area which people tend not to refer to: if you have followed Ib In Iz and you are using a standard protective device then I2 (and so 433-02-01 (iii)) doesnt really ever come into it!

You are either now happy

or in a worse state of confusion than when you started

and my typing hand is hurting

have fun!!!

Hi I think your wrong - you suddenly start referring to Ia - this is what is 5 xs the current rating
of a protective device - that is for fault or short circuit current and ionstantaneous tripping

were talking about I2 Overcurrent not fault current Ia

this current is mch smaller then ia - like a fusing factor - what the gentelmen wants to know is where is the official I2 table for Protective devices
Thanks

 

[Guest123]

OP is 2 and a half YEARS old.

 

[GStueyXR]

OP is 2 and a half YEARS old.

LOL, wonder if he got his answer yet

 

[pevvers]

Well, even though this post is 30 months old, the information can be found in the IET book on design calculations...

 

[Sparkystalbans]

Luke, you can look at this a few diferent ways. As Alan said, you can look at the graphs in the regs, or if you want the instanteneous trip current for cicrcuit breaker type, go to table 7.2b, page 51B

it depends how you phrase the question:-

If you said, how much current does it take for a 32A Type B BS 60898 circuit breaker to operate in a given time (say 0.4S), then you would be on page 199 in the regs. (or how long will it take to operate with a set/calculated/given fault current)

Wheras if you wanted to know the 'trip current' for the same MCB, you would be in table 7.2B This would tell you that a type b has an instantaneous trip current of 3 to 5 times In, so it would take 90-150A to trip the MCB

so i suppose it depends if you are foccussing on time are actual current levels.

To be technically correct, a DEVICE does not have an I2, a CIRCUIT has an I2

I2 is the 'current causing effective operation of the protective device', therefore the I2 value will be dependent on the circuit charecteristics (Zs etc), this will be co-ordinated with the In of the protective device but the two are not the same.

Have a look at 433-02-01 which may help! (or not!)

regards

Tony

Tony, it's a few years later, just reading this but thank you for such a nice explanation