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Topic Title: Meeting disconnection times for short circuit protection
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Created On: 04 December 2013 07:56 PM
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 04 December 2013 07:56 PM
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earthing

Posts: 56
Joined: 17 December 2008

How do you meet disconnection times of 0.4 seconds for fault protection, say a short circuit?

Example: 70° Thermoplastic 2.5mm Twin & Earth, 20amp radial circuit, 56 metres long.

From table I1 OSG the R1+RN (mV/A/m) value is 14.82

From table I3 OSG 70° Thermoplastic (Incorporated in a bunch) Ct = 1.2

From table 54.2 - K value for a conductor incorporated in a bunch with cables =115

So, 14.82 (mV/A/m) x 56m x Ct 1.2/1000 = 0.99


If = Uo/R1+R2 230/0.99 = 232.32 amps


S = 2.5 mm
K = 115
I = 232.32 amps

So, t = K² x S²/ If² becomes 115² x 2.5²/232.32² =1.53 seconds Failed!


Question is what do you do from here? The only option I can see is to increase the conductor size?

Edited: 04 December 2013 at 08:14 PM by earthing
 04 December 2013 08:29 PM
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alanblaby

Posts: 346
Joined: 09 March 2012

I dont think you are taking account of the overcurrent protective device, which, for say, a 20A BS88-3 fuse would disconnect a 113A load in 0.4 seconds, a 160A load in 0.1 seconds.
 04 December 2013 08:51 PM
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OMS

Posts: 19471
Joined: 23 March 2004

So, t = K² x S²/ If² becomes 115² x 2.5²/232.32² =1.53 seconds Failed


OK - what you've done is calculate the maximum time the cable can endure the fault before disconnection and suffering thermal damage.

As Alan says - check what a fault current of 230A does to the protective device - ie how fast will it operate at that amperage - if it's less than 1.53 seconds, then the cable is thermally protected for short circuit in this case - be aware that earth faults are often imposed on circuits with reduced CPC sizes ie 2.5mm/1.5mm T&E as an example

A point to note, your application of the factor of 1.2 is incorrect for short circuit or earth fault analysis of thermal withstand, it's only applicable for calculations relating to Zs where you want worst case (ie a hot, loaded conductor) - for short circuit and earh fault analysis of thermal withstand you want to check the cable at a cold, unloaded state - so don't correct for temperature, or if you do, determine R1, R2 and Rn at ambient temperature

Regards

OMS

-------------------------
Failure is always an option
 04 December 2013 11:00 PM
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earthing

Posts: 56
Joined: 17 December 2008

Hi Alan and OMS,

I didn't mention but was working off a 20amp type 60898 mcb so for this fuse the mcb would operate in 0.1s (appx 3 says anything over 100amps will cause instantaneous operation)

So after you get your answer in time, which is the maximum time the conductor can tolerate that level of current before melting or setting fire, you then check how long it would take for the fuse to operate at that level of current. If the fuse operates before the conductor melts the circuit complies.

Thanks for pointing out the 1.2 factor OMS. I will correct my calculation.

Is there any time limit on disconnection times for short circuits?

I just checked the BGB and the 0.4s disconnection seems to be for earth faults and not short circuit faults.

If you meet the maximum ZS values does this mean you automatically meet protection against short circuit faults?

Going to do some midnight reading of the regs what joy!

Thanks for your help much appreciated.
 04 December 2013 11:38 PM
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earthing

Posts: 56
Joined: 17 December 2008

And the pennies dropped further, I should have entitled the post as: Meeting Thermal Constraints or Thermal withstand.

In my original post I asked how to meet a 0.4s disconnection time for a short circuit. That was because I thought BS7671 asked for a 0.4 disconnection time for short circuit currents as well as earth faults but I can't find anything which says that is the case.

But I did find regulation 434.5.2 which says:

A fault occuring at any point in a circuit shall be interrupted within a time such that the fault current does not cause the permitted limiting temperature of any conductor or cable to be exceeded.

For a fault of very short duration (less than 0.1s) for fault current limiting devices K² S² shall be greater than the value of let through energy (I²t) quoted for the class of protective device to BS EN 60898-1, BS EN 60898-2 or BS EN 61009-1, or as quoted by the manufacturer.

The time, t, in which a given fault current will raise the live conductors from the highest permissible temperature in normal duty to the limiting temperature, can, as an approximation, be calculated from the formula:

t = K² x S² / I²

Edited: 04 December 2013 at 11:48 PM by earthing
 05 December 2013 12:06 AM
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earthing

Posts: 56
Joined: 17 December 2008

Hi OMS I run the calcs again without the 1.2 factor.

The result was that there was more fault current which led to the conductor reaching its maximum operating temperature quicker, in 1.05 seconds instead of 1.5 s.

Even though appendix 3 states that anything over 100 amps causes instantaneous disconnection, I guess this would still mean that in reality the fuse would operate quicker?

Being that regs usually errs on the side of caution would it be more desirable for the fault to disconnect faster?

I seem to remember the tutor saying something along the lines of the faster a conductor temperature raises the more damage is caused by the rapid heat rise.

Or is it more a question of accuracy? Just being more accurate with whats actually happening?
 05 December 2013 09:17 AM
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Parsley

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It's why you need to be careful using when T&E with reduced CPC's in commercial installations where the fault currents are high. You need manufacturers energy let through data, but like the vd question this isn't going to be required for 2396.

Regards
 05 December 2013 09:51 AM
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AJJewsbury

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Being that regs usually errs on the side of caution would it be more desirable for the fault to disconnect faster?

I seem to remember the tutor saying something along the lines of the faster a conductor temperature raises the more damage is caused by the rapid heat rise.

Ideally, what you want to do is minimise the total amount of energy that the cable has to absorb - as it's the energy that creates the heat that does the damage - and the amount of energy depends on the current and the time that the current flows for - if fact it's proportional to the square of the current - hence I²t. The time, starts when the fault occurs and finishes when the protective device stops the current flowing.

What's not always obvious is the relationship between disconnection time and current for different types of protective devices. For traditional fuses the overall disconnection time gets faster as the current increases - so much faster that the overall energy let-through almost always reduces with increasing current (i.e. t gets smaller faster than I² gets bigger). So odd as it might first seem, you're usually better off with large fault currents. (That's why most textbooks assume that faults at the far end of the circuit are the worst-case and only use use the circuit's maximum Zs for checking conductor size etc.)

As parsley hinted at though, MCBs are a little different - although they still generally get faster with increasing current, the increase is slower (as there are mechanical bits that have to move about there are physical limits) - so there's no guarantee that t gets smaller faster than I² gets larger - indeed most MCBs (if you check manufacturer's data) have a steadily increasing energy-let through with increasing current - hence a worry with small c.p.c.s and large fault currents. (But that's usually ignored in C&G land I'm told!)

- Andy.
 05 December 2013 03:13 PM
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earthing

Posts: 56
Joined: 17 December 2008

Very interesting. I suppose its great news for a competent designer to have all these choices regarding different fusing characteristics. If you know how to utilise them that is!

Thanks,

Dan
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