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Could someone shed some light on the following please?

411.3.2.1 implies that the 5s / 0.4s disconnection times only apply to earth faults. I've not found anything else in the regs that suggests they apply to short circuits.

435.5.2 says faults should be disconnected before the fault current causes the conductor limiting temperature to be exceeded. It goes on to give an equation to calculate the approx time that a conductor would reach its temperature limit.

Can I use this equation to find a maximum short circuit current for a circuit protected by, say, a BS 60898 MCB?
 
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Charlie_

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Use the characteristics curves to find out disconnection times/current.
If your cables are correctly selected for ccc and satisfy all your vd calculations then you will have no concerns over the cable overheating under sc conditions.
This is one of the reasons why I always test Rn as well as R1,R2.
RCBOs are being used to get round high Zs issues without any concerns for overload conditions
 

Ian1981

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Could someone shed some light on the following please?

411.3.2.1 implies that the 5s / 0.4s disconnection times only apply to earth faults. I've not found anything else in the regs that suggests they apply to short circuits.

435.5.2 says faults should be disconnected before the fault current causes the conductor limiting temperature to be exceeded. It goes on to give an equation to calculate the approx time that a conductor would reach its temperature limit.

Can I use this equation to find a maximum short circuit current for a circuit protected by, say, a BS 60898 MCB?
In a word yes.
Generally if a circuit satisfies the requirements for ADS for earth faults then it’s generally accepted that short circuit fault conditions for disconnection times are satisfied, especially with circuits with a reduced cpc.
The formula T= K2S2/I2 will give the maximum time that the live conductors can be exposed to a fault current before damage to the conductors and surrounding insulation will occur
As a 32 amp bs60898 type B will disconnect within 0.1-5 seconds at 160 amps then you will also achieve short circuit protection
 
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Thanks for taking the time to reply guys. It's a theoretical question really but I'm interested in why things are so.

If I was to use a 1mm2 circuit on a model 6A type B as an example. 30A fault or above the magnetic trip kicks in and disconnects in 0.1 second. If i rearrange the formula to

I=sqrt(K2S2/T),

and use T = 0.09second (slightly faster than MCB tripping time), and K=115, then I=383.3A.

Put that into R=V/I = 230/383.3 = 0.6 ohm minimum L-N loop impedance to avoid reaching limiting temp in the event of a short circuit?
Now, there must be a flaw in my logic somewhere, as I've never been told/read anywhere to watch out for minimum loop impedance in final circuits.
 

TJ Anderson

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Youre dead right :). It is very true, there is a minimum Zs too. In reality though it is not as it seems. BS7671 only shows 0.1s as the fastest disconnection time for 60898 but this isn't the case. As you say the B6 will disconnect in 0.1s at 30A, but with a 383A fault current would be much quicker. It can go down to 0.01s, which is realistically as fast as it could mechanically disconnect. If you had situation you are showing, albeit theoretical, then you need the manufacturers data for I2t and let through energy.
 
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Thanks TJ. I'm just looking at a Hager type B fault curve, and as you say a 5X fault current will disconnect in about .015s, with just over 6X disconnecting in .01s. There's no data for faster than than. So in reality, in the event of a short circuit, will an MCB always disconnect faster than the limiting temp of the conductor is reached?
 
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Moving the goal posts a bit lot:

In my OP, I mentioned that 411.3.2.1 seems to imply that the 5s / 0.4s disconnection times only apply to earth faults, and not short circuit faults. This appears to be backed up by data in table 7.1(i) of the OSG.

For example, take a radial in 1.0/1.0 on a 6A type D, TN-C-S. Without RCD, the circuit length is limited to 33m due to max Zs, but with RCD, the circuit is longer at 56m, limited by voltage drop. If the disconnection times did apply to short circuits, then I'd expect the limit for with RCD to again be 33m, this time limited by SC. (I would expect Zs and L-N loop impedance to be pretty much the same in this case).

Am I thinking this out right, or is there something I'm missing?

If I am right, then how do you calculate a max loop impedance? The notes to table 7.1(i) say "sc Limited by line to neutral loop impedance (short-circuit)", so it must be possible.
 

TJ Anderson

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Thanks TJ. I'm just looking at a Hager type B fault curve, and as you say a 5X fault current will disconnect in about .015s, with just over 6X disconnecting in .01s. There's no data for faster than than. So in reality, in the event of a short circuit, will an MCB always disconnect faster than the limiting temp of the conductor is reached?
Yes, it will disconnect fast enough to limit it, but as you have seen, requires manufacturer data.
In fact, staying with the example you gave, and then going with a theoretical fault current of 6kA, (the max Ipf of a typical domestic 60898), it would still comply with 1mm with regard to thermal constraints as disconnection so quick and energy let through subsequently small.

Not sure how to write small 2's on here lol....

Sq.rt 6000(sq) x 0.01 (sq) / 115 = 0.52 (min csa)
 

TJ Anderson

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Moving the goal posts a bit lot:

In my OP, I mentioned that 411.3.2.1 seems to imply that the 5s / 0.4s disconnection times only apply to earth faults, and not short circuit faults. This appears to be backed up by data in table 7.1(i) of the OSG.

For example, take a radial in 1.0/1.0 on a 6A type D, TN-C-S. Without RCD, the circuit length is limited to 33m due to max Zs, but with RCD, the circuit is longer at 56m, limited by voltage drop. If the disconnection times did apply to short circuits, then I'd expect the limit for with RCD to again be 33m, this time limited by SC. (I would expect Zs and L-N loop impedance to be pretty much the same in this case).

Am I thinking this out right, or is there something I'm missing?

If I am right, then how do you calculate a max loop impedance? The notes to table 7.1(i) say "sc Limited by line to neutral loop impedance (short-circuit)", so it must be possible.

As you say, The 0.4s and 5s disconnection times only apply to earth faults.

If ADS for earth fault is not being achieved by OCPD, but by RCD, then yes, it needs to be checked that the cable would not be overheated under short circuit conditions.

Assuming VD was ok, then you would move on to this....

I use Zn as a ref for L-N impedance recorded at furthest point in circuit....

Uo X Cmin / Zn =

This will give you a fault current to calc on adiabatic to confirm conductor size in large enough.
 
D

Deleted member 26818

Remember, the CPC is generally smaller or equal to the live conductors.
If such is the case, if it can be shown, ADS can be achieved for Earth faults, it follows ADS will also be achieved for short circuits.
 

TJ Anderson

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Remember, the CPC is generally smaller or equal to the live conductors.
If such is the case, if it can be shown, ADS can be achieved for Earth faults, it follows ADS will also be achieved for short circuits.
That works when ADS for earth fault is being met by OCPD. It would then follow that thermal constraints would also be met for short circuit due to conductor being same size or greater. The scenario the OP has hypothesised is where earth fault ADS is being met by RCD.
 
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TJ, is there an error in the equation in your post #8? If adiabatic, then the time t wouldn't be squared. So:

(√(6000² x .01))/115 =5.22mm² ?

A fault current approaching that would be pretty rare I suppose, but over 1200A not so, which would give 1.04mm² - this would mean not being able to use 1mm² for lighting circuits?

Or is there something I've missed or not understood here?
 

TJ Anderson

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Sorry, my mistake,as you say it should be Sq rt I2t. I should not have squared t. In my defence......It was late, and had had a couple of beers lol

Yes, to comply with 1mm you would need a Zn of at least 0.22. Any lower would exceed 1150A which would exceed 1mm by adiabatic.

Uo x Cmax / 1150 = 0.22 ohms

In many cases the Ze will see to it that the value is greater even before R1 + Rn is added, but it is possible.

I have considered this before. In fact many din rail mounted door bell transformers in consumer units should technically be wired in larger csa. If the PSSC is greater than 1150A (for a TNCS 0.22 Ze) and that transformer literally being terminated in same enclosure, then it would not satisfy thermal constraints. In reality though, it would not be a real world issue :), but we are theorising here. Would be a good laugh to pick up as a potential danger on an EICR though lol.

C2. Bell Tx circuit conductor in undersized and does not satisfy thermal constraints by adiabatic equation. Haha :)
 

Ian1981

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Sorry, my mistake,as you say it should be Sq rt I2t. I should not have squared t. In my defence......It was late, and had had a couple of beers lol

Yes, to comply with 1mm you would need a Zn of at least 0.22. Any lower would exceed 1150A which would exceed 1mm by adiabatic.

Uo x Cmax / 1150 = 0.22 ohms

In many cases the Ze will see to it that the value is greater even before R1 + Rn is added, but it is possible.

I have considered this before. In fact many din rail mounted door bell transformers in consumer units should technically be wired in larger csa. If the PSSC is greater than 1150A (for a TNCS 0.22 Ze) and that transformer literally being terminated in same enclosure, then it would not satisfy thermal constraints. In reality though, it would not be a real world issue :), but we are theorising here. Would be a good laugh to pick up as a potential danger on an EICR though lol.

C2. Bell Tx circuit conductor in undersized and does not satisfy thermal constraints by adiabatic equation. Haha :)
You’ve completely lost me here.:)

It’s hard to imagine that a door bell tx circuit located in a Cu will not comply with the thermal constraints if protected by a 6amp mcb in an enclosure where the fault current is designed to withstand 16ka?
 
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TJ Anderson

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You’ve completely lost me here.:)

It’s hard to imagine that a door bell tx circuit located in a Cu will not comply with the thermal constraints if protected by a 6amp mcb in an enclosure where the fault current is designed to withstand 16ka?
I like these sort of threads :)

Maybe my interpretation of BS EN 61439-3 is wrong......I see it that any device or factory wiring within the enclosure is covered for 16kA Breaking Capacity. I don't see that it would cover for thermal constraints of outgoing wiring from circuit breakers.

This is from Beama:

BS EN 61439-3 Annex ZB prescribes a specific conditional test arrangement relating to the use of
an upstream BS 88-3 (formerly BS 1361 type II) 100 A fuse-link. This 16 kA conditional test verifies
the performance of the incoming device, its connections, busbar, cable links, circuit-breakers,
residual current operated circuit-breaker and any other item in the consumer unit not separately
rated at 16 kA or higher. BS EN 61439-3 requires this rated conditional short-circuit current to be
identified by the symbol Icc and BS 7671 Regulation 536.4.201 references this symbol.
 
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