Fuse I^2t let-through and cable sizing 1

[healyx]

Hi,

Is the published total let-through energy (not pre-arc)for a fuse the largest amount of energy that a fuse will let-through for a fault in the adiabatic region (0-5s)? or is it just the worst case energy that will be let-through for the worst case fault level (e.g. device rating)?

I am looking at a gG 100A (rated 80kA) fuse that has a pre-arc let through of 14000 A^2S and total let-through of 65000 A^2S.

The device takes 5s to trip at 700A (end of adiabatic period). This equates to an I^2t let-through of 2450000 A^2S. This is higher than the published total let-through, but might be valid if the published total let-though only applies to the worst case fault. Which of these two do I use for the cable short circuit temp rise equation? Would any fault current limiting occur at only 700A? Peak current isn't altered for a fault that low according to data sheet.

I don't have access to BS88 or IEC60269-2-1 where I think the conditions for totla let-through are defined. Looking on the web, I suspect total let-through for a 100A fuse assessed at 4kA fault level.

 

[7anoter4]

First of all I don't agree the adiabatic period could be 5 sec. This would be 0.1 sec. For more than 0.1 sec.the pre-arc or
total let-through I^2*t loss its signification and you have to follow the manufacturer curve.
The standard is only a guide and some limits but could not be substitute for the manufacturer curve.
I have both standards [iec 60296-1 and 60296-2-1 and what is new in the last one is only Table E in section I CH. 7.8
discrimination when circuit breaker are used. For example for 100 A fuse-link rated the I^2t=16000[min] at 4000 A Ip .
and Table H ch. 8.7.4 "Test currents and I^2.t limits for discrimination test" when for In=100 you'll get
minimum I=2300 A I^2.t= 21200 and maximum I=4000 A I^2.t=64000 A^2.sec

 

[healyx]

Thanks 7anoter4.

Not sure my question has been answered. My question is specific to how to size a cable when protected by a fuse. I'll re-phrase.

For 5s adiabatic period, I am referring to the 5 second adiabatic limit for a cable, not the fuse itself.

For the current at which the fuse takes 5s to trip, is the let-through energy (e.g. the energy the cable sees) in this instance I^2t or is it limited to the total let-through of the fuse.

(1) I would assume the energy the cable sees at 5s current is I^2t NOT the quoted total let-through energy or pre-arc energy for the fuse. Is this right?
(2) Are you saying the total let-through energy quoted for a fuse is only relevant below 0.1s? Is this standard?

The reason I am asking is that the minimum cable size for short circuit temperature rise appears to be limited by the 5s current for a cable. A 100A fuse can hold 550A for 5s. This would let through 1500 kA^2s. The total let-through during a large fault is only about 65 kA^2s. So minimum PVC cable size is sqrt(1500k)/111 = 11 mm^2 based on a 5 second overcurrent. Of course, you can't pump 100A through a through an 11 mm2 cable anyway, but I'm just trying to understand the concept.

A similar thing happens with some circuit breakers. A 20A type C BS EN 60898 breaker can hold 200A for 5 seconds (200kA^2s) (unlikely but possible - and definitely possible with a D type breaker). This requires a 4mm2 PVC cable, which is higher than the standard 2.5mm2 cable used with 20A breakers. In this case, the breaker must exhibit some fault current limiting (for which you must check manufacturer curves) to below 77kA^2s for 2.5mm2 PVC. This seems counter-intuitive but I guess you should be checking that let-through energy does not exceed this value all the way up to 5s.

 

[7anoter4]

In IEC 60296-1 ch.1.1 is stated:
"The object of this standard is to establish the characteristics of fuses or parts of fuses (fuse base,
fuse-carrier, fuse-link) in such a way that they can be replaced by other fuses or parts of
fuses having the same characteristics provided that they are interchangeable as far as their
dimensions are concerned."
The Standard is destined for the Manufacturer in order to know how to test the product [the fuse].
The Standard[see ch.5.8.2] requires from Manufacturer to provide I^2*t as function of I up to 0.1 sec. For more than
0.1 sec [see ch.5.6.1] only I=f(t) for pre-arc time[ more than 0.1].
In order to chose a fuse for a certain circuit cable another standard has to be followed. For U.S.A the NEC[NFPA 70] is one
of this. See [for instance for IEC System]:
stevenengineering.com/tech_support/PDFs/06IECBRFUSE.pdfSimilar
If you would use the adiabatic form for more than 1 sec short-circuit clearing time the cable conductor cross section calculated
could be oversized.
But if you insist so the IEEE-80 formula[37]- ch.11.3 "Conductor sizing factors"- may be employed.
For copper conductor cable Tmax=240 [XLPE ins.] Ta=90 [rated for XLPE] t=5 sec we'll get 15.7 sqr.mm/KA

 

[7anoter4]

The heat generated in a conductor : Q=I^2*t*R
where I=the pass-through current considered constant
t=the process duration
R=Ro*[1+alpha*(T-To)] where Ro is the conductor resistance measured at To temperature[usually 20 oC]

Q is a power but the heat quantity will be Q*dt for an infinite time domain.
In this short time the conductor heat will raise with c*gamma*volume*dTemp
I^2*Ro*[1+alpha*(T-To)]*dt= c*gamma*volume*dTemp if we shall neglect the outside heat dissipation.
Ro=ro*l/scu; volume=l*scu ; c=caloric capacity of the conductor material ; gamma= specific gravity of the conductor material
TCAP= c*gamma
(I^2*ro/scu^2/TCAP)*dt=dTemp/[1+alpha*(T-To)]
Integrate the equation
(I^2*ro/scu^2/TCAP)*t=1/alpha*ln((1+alpha(Tf-To))/(1+alpha(Ti-To)))
I^2*t=scu^2*TCAP/ro/alpha*ln((1+alpha(Tf-To))/(1+alpha(Ti-To)))
If ro=ohm.m instead of ohm/m*sqr.mm we have to divide the right part by 10^4
I^2*t=scu^2*TCAP/ro/alpha*ln((1+alpha(Tf-To))/(1+alpha(Ti-To)))/10^4
For copper conductor will be:
I^2*t=scu^2*3.42/1.72/0.00393*ln((1+0.00393*(Tf-To))/((1+0.00393*(Ti-To)))
I^2*t=506*scu^2*ln((1+0.00393*(Tf-To))/((1+0.00393*(Ti-To)))
If we shall divide by 0.00393 under the logarithm and take To=20 we'll get
(1+0.00393*(Tf-To))/((1+0.00393*(Ti-To)=(Tf+234.5)/(Ti+234.5)
I^2*t=506*scu^2*ln((Tf+234.5)/(Ti+234.5))
Let's say
scu=15.7 sqr.mm
Tf=250
Ti=90
I^2*t=5 A^2*sec

 

[7anoter4]

Sorry.Correction:
I^2*t=5 kA^2*SEC= 5000000 A^2*sec

 

[healyx]

Yes, I agree with all that (except 506 should be 50600 - as you have elluded to in your correction). That is the derivation for how to find k.

I guess my question is does anybody ever check the I^2t = (k*s)^2 equation when sizing a cable at the 5 second point.

I ask because if you do check it at this point, alot of the time this places a limit on the minimum cable size, more so than energy let-through at the high fault end. Traditionally, I have only ever seen people check it at the high fault end, not the 5s point.

If a cable is considered to not loose any of the fault heat up to 5 seconds then it is quite likely this should be checked because a high resistance fault is a definite possibility and can occur in the region where a 5s clearance time occurs.

The Australian Standard AS3008 does not make this clear, nor does BS7671. This applies to both fuses and circuit breakers. Can anyone advise?

 

[rcw retired EE]

IEEE Standard 242 "Recommended Practice for Protection and Coordination of Industrial and Commercial Power Systems" uses I2t curves for cable damage that extend out to about 15 seconds. The curves are based on initial temperature, maximum fault conductor temperature and no heat release through the insulation.

Coordination with these curves using fuses has not been a problem, but maybe we haven't seen enough cases.

 

[jghrist]

I guess my question is does anybody ever check the I^2t = (k*s)^2 equation when sizing a cable at the 5 second point.

If you want to check if the fuse will protect the cable at 5 seconds, you compare the fuse TCC curve with the cable damage curve (which is based on constant I²t). You don't use the I²t let-through of the fuse.

 

[healyx]

Thanks Guys,

jghrist - I guess my question is evolving. I agree that at 5 seconds you would use the TCC vs cable damage curve. The question is, what is the upper limit at which you no longer need to check the TCC vs cable damage curve. They intersect eventually, so at what point is the interection OK?

Example.

A 2.5mm2 PVC (75 - 160 deg c) cable has a thermal limit of 77 kA^2s.

A 20A BS88 fuse can hold 80A for 5s, so let-through is 32 kA^2s. Therefore up to 5s, the cable is fine.

At about 20s, the TCC and cable damage curve intersect. At 20s, the cable would definitely be in the phase where it is able to dissipate heat at the required rate, so even through the curves intersect, this is fine (I would have thought).

Back to my question, at what point in time can you ignore an intersection? 5s is the point where the I^2t = (ks)^2 equation no longer holds, so I guess it would be 5 seconds but can't find it stated anywhere in a standard. Others have suggested after 1 second is fine. I wouldn't have thought you'd need to push it out to 15 seconds, that seems extreme.

Any thoughts?

 

[healyx]

I haven't been able to find anything definitive stating that you do/don't need to check all the way up to 5s.

In Australia, a ground/earth fault is allowed to take up to 5s to clear. The ground fault current is usually calculated as the minimum fault current that can occur for a "negligible impedance" fault.

If the calculated value for a ground fault is found to take 5s to clear, the cable must be able to withstand this current for 5s. If it takes over 5 seconds then the case fails due to disconnect time being exceeded.

The question is: If the minimum ground fault current causes a disconnect at say 1s, do you need to still check up to 5s? - which would suggest a fault of higher impedance.

The Australian wiring rules AS3000 states the current to use for the withstand calculation (for an earth cable) is for a short circuit of "negligible impedance" (5.3.3.1.3). This indicates that you don't need to check up to 5s or any other limit unless that is the time at which the protective device disconnects for a "negligible impedance" fault. I'm not sure if this is the case in other international standards.

I still maintain that by not considering the whole range up to 5s, it is possible to cook a cable during a high impedance fault and I never thought that could happen....but the standard in Australia at least doesn't require you to design that out. This only holds if it does take up to 5s for a cable to start dissapating heat.

In most circumstances, Active/Neutral cables are adequately sized for operating current requirements to not be exposed to this issue. Earth/ground cables however can definitely experience this problem (given they are usually fractionally sized to active/neutrals).

If you have ever noticed insulation damage to an earth cable and can't explain it, this might be why.

BTW, this applies to both fuse and circuit breaker protection.