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Join Date: Jul 2012
Posts: 2

Maximum Let-Through Energy of Fuse Into Cable - Max at 5 Seconds Disconnect Time

07/11/2012 8:54 AM

Hi Folks

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 A2sec).

The fuse manufacturer's only seem to publish the I2 t values for very high fault currents in the current limiting range, but as you can see from above example the largest thermal let-through energy into cable is for much lower currents. As someone put it 'a long slow blow is much more damaging for a cable than a quick high current disconnection' .

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

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Commentator

Join Date: Jun 2011
Location: Hyderabad, India.
Posts: 92
Good Answers: 2
#1

Re: Maximum Let-Through Energy of Fuse Into Cable - Max at 5 Seconds Disconnect Time

07/12/2012 6:59 PM

Mostly the Fuses are used to protect correctly the semiconductor element such as power Thyristors(SCRs or Silicon Control Rectifiers) & Power Diodes(Rectifiers) where as CBs are used for protection of cables.

I2t let through Maximum energy is given by fuse manufacturers once a fuse operates due to short circuits and fuse damaged for ever. This energy is only ¼ of a half cycle sinusoidal wave while it clears the fault in 2.5msec for 50Hz system or 2msec for 60Hz system. Such a small duration of fault clearance is important requirement for semiconductor protection.

When a SCR or a Diode is operating (conducting) on AC system, it conducts for 10msec (half cycle for 50Hz) or 8.33msec (half cycle for 60Hz). Next of half cycle they do not conduct.

During conduction, SCR or Diode junction temperature goes to 1250C when their rated current is conducting and next non-conducting half cycle junction temperature come down to small value due to heat removal by heat sink attached to them.

At this stage a fault occurs and cleared by fuse in around 2msec. This let through energy (I2t) by Fuse shall be within acceptable the maximum let through energy by SCR or Diode as given by SCR or Diode manufactures. If it exceeds, then junction temperature will go very high and SCR or Diode will go into permanent damage.

However, cables are generally protected by Circuit Breakers. A CB clears a fault not less than 5 cycles that is 100msec for 50Hz or 83.3msec for 60Hz. During fault clearance by CB, cable will be heated by I2t energy. For this fault energy the cable insulation shall not be damaged. For that cable manufactures will provide curves showing Fault current in KA (in Y-axis) and duration of short circuit in Second(X-axis) for various cables sizes. Curve starting time is 0.1Second or 5cycles for 50Hz system or 6cycles for 60Hz system

If a selected cables size in mm2 does not meet the short circuit withstanding capacity then a higher size cable in mm2 will be selected.

From this it is clear that Semiconductors are to be coordinated for time of 2msec ( best protecting equipment is FUSE) where as cables are to be coordinated for 83.3msec to 100msec and their best protecting equipment is CB. Since CB is reusable whereas Fuse is not.

There is no point for fuse manufacturers to provide I2t for cables. Cables are not sensitive as compared to semiconductor for clearing time.

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