I presume you mean 16mm² rather than 16mm....

Either way, it might start a fire before breaking the circuit.

5,000 amps in 16mm2 will fry things quickly.

Hi Lyn,

I understand it would too.

I just want to know what is the time-current curve for fusing.

Thanks,

Hi Tornado,

Yes, I mean 16mm2.

I know it will start fire, explotion and damage to surrounding equipments.

I just want to know the time current curve .

Thanks,

If it's fuse wire, then it's probably not copper.

The general fuse current response times are outlined in the SAE standards and from mempory it's something like this.

100% rating forever

150% overload for 2 mins.

300% overload for 10 secs.

500% overload less than 1 sec.

Unlike a circuit breaker with faster response times.

You will of course need to know what the initial rating for that conductor is.

Hi,

Its not a fuse wire.

We are after the Pure copper wire's fusing characteristic.

Thanks,

Then something is going on that is not contained within the relevant national electrical code.

Don't know if this will help, but

The Book Reference Data for Radio Engineers 5th edition

AWG B & S #7, 0.1443 Inch diameter copper wire fuses at 561 ampere.

AWG B & S #6, 0.1620 Inch diameter copper wire fuses at 668 ampere.

I believe this is for bare copper wire in air.

Hi,

Thanks for that.

Would you know time at the perticular current ?

Thanks,

You could look up the melting point and specific heat of copper, and from that compute how many watt-seconds it takes to heat the copper that far. Your curve will come out more or less as a hyperbola like y = k/(x-x₀).

I

EC 724

You are correct. The Onderdonk equation is an empirical formula and uses assumptions that are not necessarily applicable to every situation. It is best used in situations where the current is applied in a step function so that the heat transfer to the surrounding environment is minimized due to the short rise time. When the current is ramped then heat transfer can become an issue, requiring consideration of such things as ambient conditions, horizontal or vertical orientation, presence of insulation, enclosures, etc.

Hi 7anoter4,

Thanks for the formula. It gives us approximation of the time/current.

Obviously upstream protection and current limitation should be considered in the equation.

Thanks,

If current limitation is present, then the current will be limited to the 100% full load point, and therefore the copper wire will never melt.

There is no point in having the upstream protection rating higher than the 100% full load point of this copper wire, otherwise the melting of the copper wire will take place at a lower current than the upstream protection rating, rendering the upstream protection irrelevant and superfluous.

The undercurrent of this thread is quite abstruse, as it's outside the safety envelope of national electrical standards such as British Standard 7671. Fire indemnity insurance Risk Assessors and Loss Adjusters do understand these things, and having an installation that is not installed to the relevant standard gives them an opportunity of walking away from the damage caused, laughing all the way to the bank.

Jacklyb,

I was looking at this question earlier this week! Discussed this with an electrical engineer and looked at the arc flash calculations for my system (it ended with a #12 AWG wire approx. 15-feet or 4.5-meter long tapped from much larger conductors on an 800A breaker) with a short circuit current of about 2.5 to 5 kA. That upstream breaker had a 3300A instantaneous trip and the commercial software calculated a clearing time for that breaker of about 140 sec. I protested that the Onderdonk fusing time for this wire was approx. 0.03 sec. He said that he had never heard of a person using the melting of the wire to change the characteristics of the circuit and thus change the permissible time for arc flash calculations of incident energy.

We are still trading emails regarding this situation, because all the arc flash calculating software I know of assumes a steady state in the equipment exposed to the arc flash (no allowance for vaporized or melted materials, etc.) That appears to be where we get into the subjective analysis and the extent to which an engineer or other person is willing to go beyond the very conservative results of a calculation.

--John M.

Hi Jmueller,

In fact, the purpose of the question was to investigate/study minimum load/circuit beyond which it should be low arcflash risk to be analysed and calculated for individual circuit.

i.e If a circuit has fuse rating less than 125A (400V) then the incident energy will be low enough to cause fatal injury. The exercise minimises the time to analyse low energy circuits and unnecessary inconvenience of arcflash risk where not necessary. Generally in most industry anything below 125A fuse circuits has vast network of feeders.

Good designed circuits with 16mm2 copper cables should have fuses installed at upstream of the cables to protect the cables.

The QUESTION is what other industry set as their minimum arcflash philosophy.

Thanks,