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Topic Title: Maximum let-through energy of fuse into cable
Topic Summary: Is maximum at current corresponding to a 5 seconds disconnect time
Created On: 10 July 2012 05:57 PM
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 10 July 2012 05:57 PM
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Gilbert1

Posts: 19
Joined: 06 July 2012

Hi guys (and gals)

You might be able to assist me in relation to a query I have on the maximum let-through energy of fuse versus the thermal withstand energy of cable.

Basically my query revolves around the wording in the ETCI (Irish Regs) and IEE Wiring Regulations that state that fuse I2 t let through energy is to be less than the cable k2 S2 thermal withstand capacity, for fault disconnection times up to 5 seconds

For example if you look up the time-current fuse characteristic for a 125A NH2 gG fuse and you calculate I2 t for a fault of 630A with associated fuse operating time of 5 seconds we get a higher let-through energy (1984500 A2sec) than for a fault of 1800A with a shorter fuse operating time of 0.02 seconds (64800 A2sec). Yet the fuse manufacturers state the maximum let through energy of 125000 A2sec for a fault of 20 x In = 2500A for the 125A fuse. But if the wording in the IEE Regulations is correct in that the above adiabatic formula is valid for fault durations up to 5 seconds (as they say that the equation is considered to remain adiabatic for fault current durations up to 5 seconds, i.e., does not lose heat through insulation up to 5 secs. After 5 secs some heat will dissipate through the cable insulation into the air and so formula no longer applies), then a high resistance fault of 630A for longer duration would result in a greater energy let-through into the cable than for a high fault current of 2500A for durations less than 1/4 cycle (i.e. less than 0.01 secs).

So for example if we had a 10sq.mm Copper/PVC cable wired from above 125A fuse to a load where the fuse is providing short-circuit protection only (I know its not a practical example but it serves as a example of my above query). The cable has a thermal withstand capacity of 1152 102 = 1322500 A2sec, it seems that from a thermal capacity persepective this cable will heat up and reach it's insulation limit temperature for a 630A fault and not for a 1800A fault, as for a 630A fault the let through energy by fuse into cable (1984500 A2sec) is greater than cable withstand thermal energy (1322500 A2).

I can't understand why fuse manufacturers dont publish the let through energy for current corresponding to a 5 second disconnection time rather than for very high fault current levels ??

Regards
Gilly
 10 July 2012 06:53 PM
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OMS

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I can't understand why fuse manufacturers dont publish the let through energy for current corresponding to a 5 second disconnection time rather than for very high fault current levels ??


Above one half cycle, the current limiting effect of the fuse is diminishing up to a point at 0.1 seconds where you can now predict the performance accurate enough from time current characteristics. That's why we need manufacturers data at short times (or high currents) - to allow the current limiting effect to protect cables without the cables having to be excessive in size.

Basically, a long slow blow is much more damaging than a quick high current disconnection - which I guess you know from the numbers you worked about.

20 x In will usually put you into disconnections less than 0.1 seconds - so the current limiting effect is present

Regards

OMS

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Failure is always an option
 10 July 2012 08:57 PM
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Gilbert1

Posts: 19
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Hi OMS
You already confirmed my query.... "Basically, a long slow blow is much more damaging than a quick high current disconnection" ....exactly, which is why i said that the fuse manufacturers should state the let through energy at current corresponding to a 5 second disconnection time which will be higher than at higher currents in the current limiting time of less than 0.1 second. If we sized the conductor based on the maximum let through energy stated by fuse manufacturer the cable size may be inadequate for lower fault currents (which can occur due to high overload or high resistance earth fault).
 11 July 2012 09:12 AM
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OMS

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I'm not sure of the point you are trying to make/get to

Below about 0.1 seconds, the fuse exhibits "cut off" characteristics, the waveform is quite complex and we can only rely on manufacturers data - the cut off characteristics allow us to use smaller cables safely.

Once we get beyond that point, the wave form is reasonably sinusoidal and as such we can determine I2t via time current characteristics. (ie the joule integral is easier to determine)

You would size the conductor on the actual or probable short circuit fault curent at that point in the system.

So for exampe if your 125A Fuse is a long way into a distribution system, feeding say a dist board then you will be interested in the fault level at that point and in the fault level at the end of the circuit it supplies - ie at the downstream dist bd. If you have high fault currents that will result in disconnection in less than 0.1 seconds then you evaluate K2S2 against the manufactureres I2t data. If you have low fault currents so disconnection occurs in say 3 seconds, then you evaluate K2S2 against the fault current at that time.

Alternatively, for avoidance of risk, you evaluate at the lowest probable fault current (allowing for voltage depression, conductor resistance rising with temperature etc) and use 5 seconds - that will get you a bigger cable that is "safer" and gives other benefits such as capacity for future expansion or reduces the ownership cost due to reduced I2R loss.

If we sized the conductor based on the maximum let through energy stated by fuse manufacturer the cable size may be inadequate for lower fault currents (which can occur due to high overload or high resistance earth fault).


exactly - but then you would not be evaluating the circuit for the design or installed condition.

Perhaps you might take a look at reg 434.5.2 - that should help

Regards

OMS

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Failure is always an option
 11 July 2012 10:54 AM
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david911cockburn

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Hi Gilbert,

The disconnection time for a fault of neglibile impedance to earth (or short circuit protection) is always 0.4 seconds.
5 seconds is the maximum time that a cable should be allowed to remain over-current (or over current protection), for high starting currents etc.
 11 July 2012 10:58 AM
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OMS

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Originally posted by: david911cockburn

Hi Gilbert,

The disconnection time for a fault of neglibile impedance to earth (or short circuit protection) is always 0.4 seconds.

5 seconds is the maximum time that a cable should be allowed to remain over-current (or over current protection), for high starting currents etc.


Ohh dear , ohh dear, ohh dear - you could not be more wrong David

OMS

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Failure is always an option
 11 July 2012 11:03 AM
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marclambert

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Gilbert, please please for your own sanity and safety..do not listen to David911. About anything. Ever.
Trust OMS and the many others here who will try to help and provide the best answers possible.
Regards
Marc
 11 July 2012 01:15 PM
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AJJewsbury

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The disconnection time for a fault of neglibile impedance to earth (or short circuit protection) is always 0.4 seconds.

Not so. It depends on the type of earthing arrangement and system voltage.

For 230V/400V TN systems it's 0.4s for final circuits rated 32A or less; 5s for all other circuits.

For 230V/400V TT systems it's 0.2s and 1s respectively.

Certain special locations additionally demand 30mA RCD protection which effectively reduces the disconnection time limit further (e.g. 40ms).

(Under previous versions of the regulations it was usually 0.4s for socket outlets, 5s for fixed equipment).

- Andy.
 11 July 2012 01:58 PM
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Zs

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Just for clarity chaps, the figures above are the maximum disconnection times under earth fault conditions. Big green doesn't actually stipulate maximum disconnections for short circuit conditions. The actual disconnection time would vary depending on the fault level.

(Sorry Andy )

Zs
 11 July 2012 02:10 PM
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AJJewsbury

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Big green doesn't actually stipulate maximum disconnections for short circuit conditions

other than reg 434.5.2 - although admittedly that doesn't mandate a particular number of seconds - just 'within a time such that the fault current does not cause the permitted limiting temperature ... to be exceeded'.

(Sorry Zs )
- Andy.
 11 July 2012 02:42 PM
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Gilbert1

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Hi Dave
That is not correct. Look at 411.3.2.2 and then 411.3.2.3 where it gives different disconnection times for different voltage levels and for different applications. So for a cable feeding a distribution board the max allowable disconnection time is 5 seconds for TN installations and 1 second for TT installation. The 0.4 seconds is for socket circuits and other circuits rated under 32A
 11 July 2012 02:43 PM
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OMS

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Just for clarity chaps, the figures above are the maximum disconnection times under earth fault conditions. Big green doesn't actually stipulate maximum disconnections for short circuit conditions.


other than reg 434.5.2 - although admittedly that doesn't mandate a particular number of seconds


LMFAO - sorry both

So no maximum time in seconds - just a maximum duration such that thermal damage doesn't occur - which could be significantly longer than the "assumed" 5 seconds - which probably answers the OP as to why manufactureres don't normally state the energy let through at 5 seconds.

15 all I think -

Regards

OMS

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Failure is always an option
 11 July 2012 03:02 PM
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Gilbert1

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Hi OMS
Thanks for reply.
I think my point may be that we are in aggreement or as they say singing off the same hymn sheet. But I don't think this is widely understood from reading other similar posts and from talking to others. I think there is a misconception out there that once you look up the maximum I^t value for the fuse and check that this is less than the cable k^S^ value that the cable is protected once the fault current is not greater than the max value the I^t is declared at . So for example for my 125A fuse the max I^t declared by fuse manufacturer for a fault of 20 x In (= 2500A) is 125000 A2sec, so the misconception is that once the fault level is below 2500A and the cable k^S^ is > 12500 A2sec then the cable will be fully protected, whereas the opposite may be true, i.e. for faults below 2500A the cable may indeed NOT be protected as the I^t can be much larger than 12500 A2sec for a fault with disconnection time of several seconds.
I guess my point is that I can't see why fuse manufacturers do not make it clearer and state the maximum let through energy for faults causing disconnection time of 5 seconds (5 seconds because this is the generally accepted, although possible conservative, maximum time a current can cause an I^t level of heat increase, after 5 seconds heat will dissipate through the insulation into the air)
 11 July 2012 04:42 PM
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OMS

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Originally posted by: Gilbert1

Hi OMS

Thanks for reply.

No worries Gilly - welcome to the forum BTW

I'd ignore that halfwit Cockburn - you will lose the will to live after a few pages of trying to deal with his inane ramblings


I think my point may be that we are in aggreement or as they say singing off the same hymn sheet. But I don't think this is widely understood from reading other similar posts and from talking to others.

Well, I guess that depends on who you speak to.

I think there is a misconception out there that once you look up the maximum I^t value for the fuse and check that this is less than the cable k^S^ value that the cable is protected once the fault current is not greater than the max value the I^t is declared at .


That's a bit of a dangerous misconception, but that said, fuses exhibit almost constant I2t characteristics (the time current curve on a log/log scale is nearly a straight line) so perhaps not quite as dangerous as first impressions suggest.

So for example for my 125A fuse the max I^t declared by fuse manufacturer for a fault of 20 x In (= 2500A) is 125000 A2sec, so the misconception is that once the fault level is below 2500A and the cable k^S^ is > 12500 A2sec then the cable will be fully protected, whereas the opposite may be true, i.e. for faults below 2500A the cable may indeed NOT be protected as the I^t can be much larger than 12500 A2sec for a fault with disconnection time of several seconds.

OK - lets also consider the other half of this - the cable. With a 125A fuse in place, it's likely that you'll have something like a 50mm2 cable with a 125A fuse. lets assume K= 115, so the total K2S2 = 33062500A2S.

At very short durations, the cut off characteristics of the fuse at 20 x In are 12500 A2sec so no drama.

Taking say 1 second disconnection, we need 1000A to flow and thus I2t = 1000000A2sec - again no drama

taking 5 seconds, we need 680A to flow so we have a total let through of 2312000 - again well within the cable capacity

Of course, if the fuse is only offering short circuit protection and you have much smaller cables then you are heading into dangerous territory


I guess my point is that I can't see why fuse manufacturers do not make it clearer and state the maximum let through energy for faults causing disconnection time of 5 seconds

Well they do - it's called a time current characteristic

as I said, it's only when you have very short disconnection times that you need manufactures joule integral data


(5 seconds because this is the generally accepted, although possible conservative, maximum time a current can cause an I^t level of heat increase, after 5 seconds heat will dissipate through the insulation into the air)

To be clear, some heat will be lost - compared to the inflow of energy though it's not a big percentage and your constraint as Zs and Andy pointed out above is not melting the insulation - ie you need to keep disconnection time less than that required by the equation in 434.5.2 - or use a bigger cable.


Regards

OMS

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Failure is always an option
 12 July 2012 09:12 AM
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Gilbert1

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Hi OMS
Thanks for welcome. It is an interesting Forum site.
Thanks for taking the time to go through my post and commenting.
I agree that if we select a cable that has the required CCC appropriate to the fuse size (i.e. Iz >= In / correction factors) then that cable will most likely be protected for the lower fault currents associated with disconnection times of 1 second and above, but the difficulty could arise if the fuse is providing short circuit protection only for cable (with overload protection being provided bt downstream device).

I take your point that the TCC curves providing means to calculate the I^t, but I still think it could be made clearer. I came across one fuse manufacturer, GEC Alsthom, that made this very clear and even gave the adiabatic formula I2t = k2S2 and stated "where I = current which causes fuse to operate in 5 seconds, t = 5 seconds...." Now that is clear (see http://1.234.22.71/datasheets/130475_101341.pdf).
Thanks
and thanks to you other guys that replied to me
 12 July 2012 09:18 AM
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Gilbert1

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The weblink didn't attach properly ?
Try page 8 of http://www.weutscheck.com/atta...ustrial_Fuse_Links.pdf
 12 July 2012 10:09 AM
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OMS

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Originally posted by: Gilbert1

Hi OMS

Thanks for welcome. It is an interesting Forum site.

Thanks for taking the time to go through my post and commenting.

I agree that if we select a cable that has the required CCC appropriate to the fuse size (i.e. Iz >= In / correction factors) then that cable will most likely be protected for the lower fault currents associated with disconnection times of 1 second and above, but the difficulty could arise if the fuse is providing short circuit protection only for cable (with overload protection being provided bt downstream device).

Well that's your job as the designer to first recognise the risk and then address it as part of the cabling solution I guess

I take your point that the TCC curves providing means to calculate the I^t, but I still think it could be made clearer. I came across one fuse manufacturer, GEC Alsthom, that made this very clear and even gave the adiabatic formula I2t = k2S2 and stated "where I = current which causes fuse to operate in 5 seconds, t = 5 seconds...." Now that is clear (see http://1.234.22.71/datasheets/...75_101341.pdf).


Well if you talk to fuse manufacturers they will give all sorts of lovely data for cut off characteristics, discrimination (or selectivity) analysis and data on back up protection. It may be in slightly different formats on different pages but it's all out there if you look.

I guess my point is who are we trying to make it "clearer" for ?


Thanks

and thanks to you other guys that replied to me


Regards

OMS

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 12 July 2012 10:16 AM
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AJJewsbury

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The weblink didn't attach properly ?


The site doesn't allow new users to post links (it's a defence against spam advertisers etc.). At some point you'll become "trusted" and can post links like the rest of us.

In the meantime, I think this was it: http://www.weutscheck.com/atta...strial_Fuse_Links.pdf
- Andy.
 12 July 2012 12:55 PM
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Gilbert1

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Thanks Andy thats the one, on page 8.
How long does it take to become 'trusted'.......how long is a piece of string !!!! :-)
 12 July 2012 01:40 PM
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AJJewsbury

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How long does it take to become 'trusted'

The moderators won't say! (it would then be too easy for spammers just to post the required number of posts or wait the required length of time).
- Andy.
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